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CHEMISTRY, M8 2020 HSC 1 MC

What is the function of the magnetic field in a mass spectrometer?

  1. It detects the mass of the particles.
  2. It deflects the stream of charged particles.
  3. It excites electrons to higher energy levels.
  4. It produces a stream of electrons that bombards the sample.
Show Answers Only

`B`

Show Worked Solution

Magnetic fields cause charged particles to travel in a circular path.

`=> B`

CHEMISTRY, M8 2022 HSC 28

The iron content of an impure sample (4.32 g) was determined by the process shown in the flow chart.
 

  1. Identify the brown precipitate formed at the end of step 3.   (1 mark)
  1. Calculate the percentage of iron in the original impure sample if 4.21 g of iron(`text{III}`) oxide `(\text{Fe}_2\text{O}_3)` was collected. Assume that all the iron was converted to iron(`text{III}`) oxide.   (4 marks)
Show Answers Only
  1. `text{Fe(OH)}_3`
  2. `text{68.2%}`
Show Worked Solution

a.  `text{Fe(OH)}_3`

b.   `text{n(Fe}_2 text{O}_3)` `= text{m} / text{MM}`
    `=4.21 / [2 xx55.85+3 xx 16.00]`
    `=0.0264  text{mol}`

 
`text{n(Fe)}= 2 xx text{n(Fe}_2 text{O}_3 )=0.0527  text{mol}`

`text{m(Fe)}= text{n} xx  text{MM}=0.0527238 xx 55.85=2.94\ text{g}`

`text{Fe(%)}` `= text{m(Fe)} / text{m(original sample)} xx 100%`
  `=2.944627 / 4.32 xx 100%`
  `=68.2%`

 
Therefore, the percentage of iron in the original impure sample is 68.2%.

CHEMISTRY, M8 2022 HSC 27

A bottle labelled 'propanol' contains one of two isomers of propanol.

  1. Draw the TWO isomers of propanol.   (2 marks)
     

  1. Describe how \( \ce{^13C NMR}\) spectroscopy might be used to identify which isomer is in the bottle.   (2 marks)
  1. Each isomer produces a different product when oxidised.
  2. Write equations to represent the oxidation reactions of the two isomers. Include reaction conditions.   (3 marks)
Show Answers Only

a.    Isomer 1:

Isomer 2:

b.   Identifying isomers with \( \ce{^13C NMR} \) spectroscopy:

→ this can be used to identify the isomers in the bottle because they show a different number of signals which helps deduce the carbon environment.

→ Propan-1-ol contains 3 \( \ce{C}\) environments so it would have 3 peaks on a \( \ce{^13C NMR}\) spectrum whereas propan-2-ol only contains 2 \( \ce{C}\) environments (due to symmetry), so it would only have 2 signals on a \( \ce{^13C NMR}\) spectrum.

c.  


  

Show Worked Solution

a.    Isomer 1:

   

Isomer 2:

   

b.   Identifying isomers with \( \ce{^13C NMR} \) spectroscopy:

→ this can be used to identify the isomers in the bottle because they show a different number of signals which helps deduce the carbon environment.

→ Propan-1-ol contains 3 \( \ce{C}\) environments so it would have 3 peaks on a \( \ce{^13C NMR}\) spectrum whereas propan-2-ol only contains 2 \( \ce{C}\) environments (due to symmetry), so it would only have 2 signals on a \( \ce{^13C NMR}\) spectrum.

c.  


  


♦ Mean mark (c) 52%.

CHEMISTRY, M6 2022 HSC 26

Students conducted an experiment to determine `Delta H` for the reaction between sodium hydroxide and hydrochloric acid.

The data from one student are shown in the table below.
 

Assume that all the solutions have the same specific heat capacity as water.

  1. Calculate the heat energy released in this experiment.   (2 marks)
  1. A second student using the same procedure obtained `2.6 × 10^(3)\ text{J}` for the heat energy released in their experiment.
  2. Use this value to determine the enthalpy of neutralisation, `Delta H`, in `\text{kJ}\ text{mol}^(-1)`, for the reaction shown.
  3.      \( \ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)} \)   (2 marks)
Show Answers Only
  1. `2800  text{J}`
  2. `-52\ text{kJ mol}^-1`
Show Worked Solution

a.   `T_(text{avg initial}) = (21.0 + 21.2) / 2 = 21.1 text{°C}`

`q` `= mc DeltaT`
  `= (100.7 + 102.0) (4.18) (24.4 − 21.1)`
  `=2796.04…`
  `= 2800  text{J (to 2 s.f.)}`

 
Therefore, 2800 J of heat energy is released in this experiment (assuming no energy loss).
 

b.   `q = 2600\ text{J}`

`text{n} = 0.1 xx 0.5 = 0.05\ text{mol}`

`DeltaH` `= -q/text{n}`
  `= -2600 / 0.05`
  `= -52\ 000\ text{J mol}^-1`
  `= -52\ text{kJ mol}^-1`

♦ Mean mark (b) 43%.

CHEMISTRY, M6 2022 HSC 25

The pH of two aqueous solutions was compared.
 

Explain why the `\text{HCN}(aq)` solution has a higher pH than the `\text{HCl}(aq)` solution. Include a relevant chemical equation for the `\text{HCN}(aq)` solution.   (3 marks)

Show Answers Only

→ \( \ce{HCl} \) is a strong acid, ie it completely ionises in water to form \( \ce{H+} \) ions.

→ On the other hand, \( \ce{HCN} \) is a weak acid, ie it partially ionises in water to form \( \ce{H+} \) ions.

\( \ce{HCl(aq) -> H+ (aq) + Cl- (aq)} \)

\( \ce{HCN(aq) \rightleftharpoons H+ (aq) + CN– (aq)}\)

→ As \( \ce{[H+]} \) decreases, pH increases  (\( \ce{\text{pH} = – log [H+]} \))

→ Therefore, at the same 0.2M, the \( \ce{HCN} \) solution would have a lower \( \ce{[H+]} \) and thus would have a higher pH than \( \ce{HCl} \).

Show Worked Solution

→ \( \ce{HCl} \) is a strong acid, ie it completely ionises in water to form \( \ce{H+} \) ions.

→ On the other hand, \( \ce{HCN} \) is a weak acid, ie it partially ionises in water to form \( \ce{H+} \) ions.

\( \ce{HCl(aq) -> H+ (aq) + Cl- (aq)} \)

\( \ce{HCN(aq) \rightleftharpoons H+ (aq) + CN– (aq)}\)

→ As \( \ce{[H+]} \) decreases, pH increases  (\( \ce{\text{pH} = – log [H+]} \))

→ Therefore, at the same 0.2M, the \( \ce{HCN} \) solution would have a lower \( \ce{[H+]} \) and thus would have a higher pH than \( \ce{HCl} \).


Mean mark 57%.

CHEMISTRY, M5 2022 HSC 23

Consider the following system which is at equilibrium in a rigid, sealed container.

\( \ce{4NH3(g) + 5O2(g) \rightleftharpoons 4NO(g) + 6H2O(g)} \ \ \ \ \ \  \Delta H = -950\ \text{kJ mol}^{-1} \) 

  1. Identify what would happen to the amount of \( \ce{NO(g)} \) if the temperature was increased.   (1 mark)
  1. Explain why a catalyst does not affect the equilibrium position of this system.   (2 marks)
  1. Using collision theory, explain what would happen to the concentration of \( \ce{NO(g)} \) if \( \ce{H2O(g)} \) was removed from the system.   (3 marks)
Show Answers Only

a.   The amount of \( \ce{NO2}\) decreases.

b.   Catalyst affect on equilibrium:

→ A catalyst lowers the activation energy and equally increases the rate of both the forward and reverse reactions.

→ As a result, there is no net change in equilibrium, and thus the equilibrium position remains unchanged.

c.   If \( \ce{H2O}\) is removed from the system:

→ This would cause the reverse reaction to decrease because there would be a lower likelihood of successful collisions between \( \ce{NO}\) and \( \ce{H2O}\) molecules.

→ As a result, the forward reaction rate is greater than the reverse reaction rate, thus, the equilibrium would shift to the right, causing \( \ce{[NO]}\) and \( \ce{[H2O]}\) to increase.

→ As the equilibrium is shifting to the right, the forward reaction rate decreases, whilst the reverse reaction rate increases, until they reach equilibrium.

→ At equilibrium, the concentration of all substances remain constant.

Show Worked Solution

a.   The amount of \( \ce{NO2}\) decreases.
 

b.   Catalyst affect on equilibrium:

→ A catalyst lowers the activation energy and equally increases the rate of both the forward and reverse reactions.

→ As a result, there is no net change in equilibrium, and thus the equilibrium position remains unchanged.
 

c.   If \( \ce{H2O}\) is removed from the system:

→ This would cause the reverse reaction to decrease because there would be a lower likelihood of successful collisions between \( \ce{NO}\) and \( \ce{H2O}\) molecules.

→ As a result, the forward reaction rate is greater than the reverse reaction rate, thus, the equilibrium would shift to the right, causing \( \ce{[NO]}\) and \( \ce{[H2O]}\) to increase.

→ As the equilibrium is shifting to the right, the forward reaction rate decreases, whilst the reverse reaction rate increases, until they reach equilibrium.

→ At equilibrium, the concentration of all substances remain constant.

CHEMISTRY, M6 2022 HSC 15 MC

A 25.00 mL sample of 0.1131 `\text{mol}\ text{L}^(-1)` `\text{HCl}(aq)` was titrated with an aqueous ammonia solution. The conductivity of the solution was measured throughout the titration and the results graphed.
 

What was the concentration of the ammonia solution?

  1. 0.0452 `\text{mol}\ text{L}^(-1)`
  2. 0.189 `\text{mol}\ text{L}^(-1)`
  3. 0.283 `\text{mol}\ text{L}^(-1)`
  4. 0.690 `\text{mol}\ text{L}^(-1)`
Show Answers Only

`C`

Show Worked Solution

The equivalence point occurs at 10.0 mL of ammonia.

`text{HCl} (aq) + text{NH}_3 (aq) → text{NH}_4 text{Cl} (aq)`

`text{n(HCl)}` `= text{c} xx text{V}`
  `= 0.1131 xx 0.02500`
  `= 2.828 xx  text{10}^(−3)  text{mol}`

 
`text{n(NH}_3 )= text{n(HCl)}= 2.828 xx text{10}^(−3)  text{mol}`

`:.\ text{[NH}_3]` `= text{n} / text{V}`
  `= [2.828 xx text{10}^(−3)] / 0.0100`
  `= 0.283  text{mol L}^(–1)`

 
`=> C`

CHEMISTRY, M5 2022 HSC 13 MC

Nitrosyl bromide decomposes according to the following equation.

\(\ce{2NOBr(g)  \rightleftharpoons  2NO(g) + Br2(g)}\)

A 0.64 mol sample of \(\ce{NOBr}\) is placed in an evacuated 1.00 L flask. After the system comes to equilibrium, the flask contains 0.46 mol \(\ce{NOBr}\).

What are the concentrations of \(\ce{NO}\) and \(\ce{Br2}\) in the flask at equilibrium?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1.2ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1.2ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1.2ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1.2ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline\ce{[NO]}\left( \text{mol L}^{-1}\right) &  \ce{\left[Br_2\right]} \left(\text{mol L}^{-1}\right)\\
\hline \rule{0pt}{2.5ex}0.18 \rule[-1ex]{0pt}{0pt}& 0.09 \\
\hline \rule{0pt}{2.5ex}0.18 \rule[-1ex]{0pt}{0pt}& 0.18 \\
\hline \rule{0pt}{2.5ex}0.36 \rule[-1ex]{0pt}{0pt}& 0.18 \\
\hline \rule{0pt}{2.5ex}0.92 \rule[-1ex]{0pt}{0pt}& 0.46 \\
\hline
\end{array}
\end{align*}

Show Answers Only

`A`

Show Worked Solution

\begin{array} {|l|c|c|c|}
\hline  & \text{NOBr} & \text{NO} & \text{Br}_2 \\
\hline \text{Initial} & 0.64 & 0 & 0 \\
\hline \text{Change} & -0.18 & +0.18 & +0.09 \\
\hline \text{Equilibrium} & 0.46 & 0.18 & 0.09 \\
\hline \end{array}

 
`[text{NO}] = 0.18\ text{mol L}^-1`

`[text{Br}_2] = 0.09\ text{mol L}^-1`

`=> A`

CHEMISTRY, M7 2022 HSC 11 MC

Cyclohexanol is an alcohol and undergoes a dehydration reaction with sulfuric acid as shown.
 

What is the major organic product of this reaction?
 

 

Show Answers Only

`B`

Show Worked Solution

When alcohols are dehydrated using concentrated \(\ce{H2SO4}\), an \(\ce{OH}\) group and a \(\ce{H}\) atom from the adjacent carbon is eliminated to form an alkene.

`=> B`

Complex Numbers, EXT2 N2 2022 HSC 13c

Consider the equation  `z^5+1=0`, where `z` is a complex number.

  1. Solve the equation  `z^5+1=0`  by finding the 5th roots of `-1`.  (2 marks)

  2. Show that if `z` is a solution of  `z^5+1=0`  and  `z !=-1`, then  `u=z+(1)/(z)`  is a solution of  `u^2-u-1=0`.  (2 marks)

  3. Hence find the exact value of  `cos\ (3pi)/(5)`.  (3 marks)

Show Answers Only
  1. `z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`
  2. `text{Proof (See Worked Solutions)}`
  3. `(1-sqrt5)/4`
Show Worked Solution

i.   `z^5+1=0\ \ =>\ \ z^5=-1`

`z=e^(i((2k+1)/5)),\ \ kin{0,1,-1,2,-2}`

`:.z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`
  

ii.   `z^5+1=(z+1)(z^4-z^3+z^2-z+1)`

`text{Given}\ \ z!=-1,`

`z^4-z^3+z^2-z+1=0`
 

`text{Divide by}\ z^2\ \ (z!=0)`

`z^2-z+1-1/z+1/z^2` `=0`  
`z^2+1/z^2-(z+1/z)+1` `=0`  
`z^2+2+1/z^2-(z+1/z)-1` `=0`  
`(z+1/z)^2-(z-1/z)-1` `=0`  

 
`text{Let}\ \ u=z+1/z:`

`:.u^2-u-1=0`
 


Mean mark (ii) 53%.

iii.  `u^2-u-1=0`

`text{By quadratic formula:}`

`u` `=(1+-sqrt(1-4xx1xx(-1)))/(2)`  
  `=(1+-sqrt5)/2`  

♦ Mean mark (iii) 43%.
`z+1/z` `=(1+-sqrt5)/2`  
`e^(i(3pi)/5)+e^(-i(3pi)/5)` `=(1-sqrt5)/2,\ \ (cos\ (3pi)/5 <0)`  
`2cos((3pi)/5)` `=(1-sqrt5)/2`  
`cos((3pi)/5)` `=(1-sqrt5)/4`  

Proof, EXT2 P1 2022 HSC 13a

Prove that for all integers `n` with `n >= 3`, if `2^(n)-1` is prime, then `n` cannot be even.  (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text{Contrapositive statement:}`

`text{If}\ n\ text{is even}, 2^n-1\ text{is NOT prime.}`

`text{Let}\ \ n=2k,\ \ (kinZZ and k>=2)`

`2^n-1` `=2^(2k)-1`  
  `=(2^k)^2-1`  
  `=(2^k-1)(2^k+1)`  

 
`text{S}text{ince}\ \ k>=2\ \ =>\ \ 2^k-1>=3 and 2^k+1>=5`

`:.2^n-1\ \ text{is not prime if}\ n\ text{is even, as it has two non-trivial integer factors.}`

`:.\ text{By contrapositive statement, if}\ 2^(n)-1\ text{is prime}, n\ text{cannot be even.}`

Complex Numbers, EXT2 N1 2022 HSC 12e

Given the complex number  `z=e^(i theta)`, show that  `w=(z^(2)-1)/(z^(2)+1)`  is purely imaginary.  (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`z=e^(i theta)`

`w` `=(z^2-1)/(z^2+1)`  
  `=(e^(i2theta)-1)/(e^(i2theta)+1)`  
  `=(e^(i2theta)-1)/(e^(i2theta)+1) xx (e^(-i2theta)+1)/(e^(-i2theta)+1)`   
  `=(1+e^(i2theta)-e^(-i2theta)-1)/(1+e^(i2theta)+e^(-i2theta)+1)`  
  `=(e^(i2theta)-e^(-i2theta))/(2+e^(i2theta)+e^(-i2theta))`  
  `=(cos(2theta)+isin(2theta)-(cos(-2theta)+isin(-2theta)))/(2+cos(2theta)+isin(2theta)+cos(-2theta)+isin(-2theta))`  
  `=(cos(2theta)+isin(2theta)-cos(2theta)+isin(2theta))/(2+cos(2theta)+isin(2theta)+cos(2theta)-isin(2theta))`  
  `=(i2sin(2theta))/(2+2cos(2theta))`  
  `=i((2sin(2theta))/(2+2cos(2theta)))\ \ text{(purely imaginary)}`  

CHEMISTRY, M6 2022 HSC 10 MC

Which equation shows the hydrogen carbonate ion acting as a Brønsted-Lowry acid?

  1. \(\ce{HCO3-(aq) \rightleftharpoons CO3^2-(aq) + H+(aq)}\)
  2. \(\ce{HCO3-(aq) + H2O(l) \rightleftharpoons H2CO3(aq) + OH-(aq)}\)
  3. \(\ce{HCO3-(aq) + NH3(aq) \rightleftharpoons CO3^2-(aq) + NH4+(aq)}\)
  4. \(\ce{HCO3-(aq) + HCOOH(aq) \rightleftharpoons HCOO-(aq) + H2CO3(aq)}\)
Show Answers Only

`C`

Show Worked Solution

\(\ce{HCO3-}\) acts as a Bronsted-Lowry acid and donates a proton to \(\ce{NH3}\).

`=> C`

CHEMISTRY, M5 2022 HSC 8 MC

A system is described as follows.

\(\ce{2NaHCO3(s)  \rightleftharpoons Na2CO3(s) + CO2(g) + H2O(g)}\)

What is the equilibrium expression for this system?

  1. `K_{eq}=[\text{CO}_2]`
  2. `K_{eq}=[\text{CO}_2][\text{H}_2\text{O}]`
  3. `K_{eq}=(1)/([\text{CO}_2][\text{H}_2\text{O}])`
  4. `K_{eq}=([\text{Na}_2\text{CO}_3][\text{CO}_2][\text{H}_2\text{O}])/[\text{NaHCO}_3]^(2)`
Show Answers Only

`B`

Show Worked Solution

→ The concentrations of solids and pure liquids are omitted from the equilibrium expression because they have a constant concentration.

→ Thus, the equilibrium expression is:

`K_(eq) = [text{CO}_2] [text{H}_2 text{O}]`

`=> B`

CHEMISTRY, M8 2022 HSC 4 MC

An analytical chemist was using atomic absorption spectroscopy (AAS) to determine the manganese concentration in a sample.

The following diagram shows the absorbance lines of manganese.
 

The diagrams below show the emission spectra of four AAS lamps.

Which lamp should be used to determine the manganese concentration in the sample?
 

 

Show Answers Only

`C`

Show Worked Solution

Consider Option C:

→ The emission spectrum of this lamp emits wavelengths of light that correspond to the absorption spectrum of `text{Mn}`.

→ This lamp would be best used to determine the `text{Mn}` concentration because it will emit wavelengths of light that are only absorbed by `text{Mn}`.

`=> C`

CHEMISTRY, M5 2021 HSC 33

The relationships between `Delta H` and `TDelta S` with temperature for a chemical system are displayed in the graph.
 
   

  1. Calculate `Delta G` for this system at 300 K.   (2 marks)

  2. What can be deduced about the system when the temperature is `T_1`, `T_2` and `T_3`? Support your answer with reference to the graph.  (4 marks)

Show Answers Only

a.  `-15\ text{kJ/mol}`

b.   → For all three reactions `ΔH < 0` (all are exothermic).

→ The entropy of the reaction, `ΔS` is negative as `TΔS` is negative (`T>0`).

→ From the relationship `ΔG = ΔH − TΔS`, we can deduce whether the reaction will be spontaneous (`ΔG<0`) or non-spontaneous (`ΔG>0`). 

→ At `text{T}_1`:  `ΔH` is more negative than `– TΔS`, and so `ΔG<0`. Thus the reaction is spontaneous.

→ At `text{T}_2`: `ΔH` is equal to `– TΔS`, and so `ΔG=0`. Thus the system is in equilibrium.

→ At `text{T}_3`: `ΔH` is less negative than `– TΔS`, and so `ΔG<0`. Thus the reaction is non-spontaneous.

Show Worked Solution

a.   At Temperature = 300K:

`TΔS = – 78\ text{kJ/mol},  ΔH = –93\ text{kJ/mol}`

`ΔG` `=ΔH-TΔS `  
  `=-93-(-78)`  
  `=-15\ text{kJ/mol}`  

 
b.   → For all three reactions `ΔH < 0` (all are exothermic).

→ The entropy of the reaction, `ΔS` is negative as `TΔS` is negative (`T>0`).

→ From the relationship `ΔG = ΔH − TΔS`, we can deduce whether the reaction will be spontaneous (`ΔG<0`) or non-spontaneous (`ΔG>0`). 

→ At `text{T}_1`:  `ΔH` is more negative than `– TΔS`, and so `ΔG<0`. Thus the reaction is spontaneous.

→ At `text{T}_2`: `ΔH` is equal to `– TΔS`, and so `ΔG=0`. Thus the system is in equilibrium.

→ At `text{T}_3`: `ΔH` is less negative than `– TΔS`, and so `ΔG<0`. Thus the reaction is non-spontaneous.


Mean mark (b) 55%.

CHEMISTRY, M8 2021 HSC 30

A student was trying to identify the ions present in a dilute aqueous solution.

The solution contained ions of barium, calcium or magnesium, and ions of hydroxide or acetate.

The student performed the following tests and recorded their observations. A fresh sample of the solution was used for each test.

    • When aqueous sodium chloride was added, no visible reaction was observed.
    • When aqueous silver nitrate was added, brown precipitate was produced. The precipitate dissolved when dilute hydrochloric acid was added.
    • When concentrated aqueous sodium sulfate was added, white precipitate was produced.

Evaluate this procedure as a method of identifying the ions.   (5 marks)

Show Answers Only

→ The first test is unnecessary for the student to identify cations or anions.

→ All possible ions present will not form a precipitate with the `text{NaCl}` reagent being added.
 

→ The second test is an appropriate method to identify the presence of cations or anions.

→ The addition of silver nitrate will form a precipitate with hydroxide ions but not acetate ions. Thus, the brown precipate formed by the second test indicates the presence of hydroxide ions.

→ Additionally, it also eliminates the presence of magnesium cation, as magnesium hydroxide is insoluble.
 

→ The third test is an invalid method.

→ Sodium sulfate will form a precipitate with both barium and calcium ions, giving both a white precipitate. Thus, it is not possible to accurately distinguish between calcium and barium ions.

→ Instead, a flame test would be best to identify the cations, where a red flame will be found for calcium and a green flame would be found for barium.

Show Worked Solution

→ The first test is unnecessary for the student to identify cations or anions.

→ All possible ions present will not form a precipitate with the `text{NaCl}` reagent being added.
 

→ The second test is an appropriate method to identify the presence of cations or anions.

→ The addition of silver nitrate will form a precipitate with hydroxide ions but not acetate ions. Thus, the brown precipate formed by the second test indicates the presence of hydroxide ions.

→ Additionally, it also eliminates the presence of magnesium cation, as magnesium hydroxide is insoluble.
 

→ The third test is an invalid method.

→ Sodium sulfate will form a precipitate with both barium and calcium ions, giving both a white precipitate. Thus, it is not possible to accurately distinguish between calcium and barium ions.

→ Instead, a flame test would be best to identify the cations, where a red flame will be found for calcium and a green flame would be found for barium.


♦ Mean mark 47%.

CHEMISTRY, M7 2021 HSC 26

A sequence of chemical reactions, starting with 2-methylprop-1-ene, is shown in the flow chart.

  1. Complete the flow chart by drawing structural formulae for compounds `text{A}`, `text{B}`, `text{C}`, and `text{D}`.   (4 marks)
     
     

     
  2. Reflux is used in the synthesis of methyl 2-methylpropanoate.
  3. Provide TWO reasons for using this technique.   (2 marks)

Show Answers Only

a.   Compound A:

Compound B:

Compound C:

Compound D:


 

b.  Reasons for reflux technique:

→ Reflux heats the reaction mixture which increases the average kinetic energy, and thus increases the reaction rate.

→ Heating causes the volatile substances to form vapour molecules. Refluxing uses a condenser to cool the vapour molecules into liquids, and thus retains the substances.

Show Worked Solution

a.   Compound A:

Compound B:

Compound C:

Compound D:


 

b.  Reasons for reflux technique:

→ Reflux heats the reaction mixture which increases the average kinetic energy, and thus increases the reaction rate.

→ Heating causes the volatile substances to form vapour molecules. Refluxing uses a condenser to cool the vapour molecules into liquids, and thus retains the substances.


♦ Mean mark (b) 46%.

CHEMISTRY, M7 2021 HSC 24

A straight-chained alkane has a molar mass of 72.146 g mol ¯1.

Provide the structural formulae for this alkane and all of its isomers.

Name these molecules using IUPAC conventions.   (4 marks)

Show Answers Only

Straight chained alkane (Pentane):

2-methylbutane:

2,2-dimethylpropane:

Show Worked Solution

Straight chained alkane (Pentane):

2-methylbutane:

2,2-dimethylpropane:

CHEMISTRY, M6 2021 HSC 23

Methanoic acid reacts with aqueous potassium hydroxide. A salt is produced in this reaction.

  1. Write a balanced chemical equation for this reaction.   (2 marks)

  2. Is the salt acidic, basic or neutral? Justify your answer.  (2 marks)

Show Answers Only

a.   `text{HCOOH}_((aq)) + text{KOH}_((aq))  -> text{KHCOO}_((aq)) + text{H}_2text{O}_((l))`, or

`text{HCOOH}_((aq)) + text{KOH}_((aq))  -> text{KHCOO}_((s)) + text{H}_2text{O}_((l))`

 

b.   Potassium methanoate is a basic salt.

This is because the `text{HCOO}^-` is a moderately strong conjugate base as it comes from a weak acid `text{HCOOH}`.

Show Worked Solution

a.   `text{HCOOH}_((aq)) + text{KOH}_((aq))  -> text{KHCOO}_((aq)) + text{H}_2text{O}_((l))`, or

`text{HCOOH}_((aq)) + text{KOH}_((aq))  -> text{KHCOO}_((s)) + text{H}_2text{O}_((l))`

 

b.   Potassium methanoate is a basic salt.

This is because the `text{HCOO}^-` is a moderately strong conjugate base as it comes from a weak acid `text{HCOOH}`.

CHEMISTRY, M8 2021 HSC 21

Four organic liquids are used in an experiment. The four liquids are

    • hexane
    • hex-1-ene
    • propan-1-ol
    • propanoic acid
  1. State ONE safety concern associated with organic liquids and suggest ONE way to address this safety concern.   (2 marks)

  2. The organic liquids are held separately in four flasks but the flasks are not labelled. Tests were conducted to identify these liquids. The outcomes of the tests are summarised below.   (2 marks)
     
           
    Identify the FOUR liquids.
     
           
  3. What chemical test, other than those used in part (b), could be used to confirm the identification of ONE of the liquids? Include the expected observation in your answer.   (2 marks)

Show Answers Only

a.    A safety concern is that the organic liquids are flammable.

To address this, keep substance away from open flames and keep away from ignition sources.
 

b.    Flask 1: propanoic acid  (carboxylic acids can’t be oxidised and are polar)

Flask 2: hex-1-ene (alkenes can be oxidised and are non-polar)

Flask 3: propan-1-ol (primary alcohols can be oxidised and are polar)

Flask 4: hexane (alkanes don’t react with acidified oxidants and are non-polar)
 

c.   Hex-1-ene

→ Could be identified using the bromine water test.

→ The addition of brown bromine water to an alkene causes an addition reaction where the solution changes colours from brown to colourless.

Propanoic acid

→ Could be identified through a neutralisation reaction using `text{Na}_2text{CO}_3`.

→ Effervescent reaction will result.

Propan-1-ol

→ Could be identified through an oxidation reaction using acidified dichromate.

→ The reaction would cause the solution to change from green to orange.

Show Worked Solution

a.    A safety concern is that the organic liquids are flammable.

To address this, keep substance away from open flames and keep away from ignition sources.
 

b.    Flask 1: propanoic acid  (carboxylic acids can’t be oxidised and are polar)

Flask 2: hex-1-ene (alkenes can be oxidised and are non-polar)

Flask 3: propan-1-ol (primary alcohols can be oxidised and are polar)

Flask 4: hexane (alkanes don’t react with acidified oxidants and are non-polar)
 

c.   Hex-1-ene

→ Could be identified using the bromine water test.

→ The addition of brown bromine water to an alkene causes an addition reaction where the solution changes colours from brown to colourless.

Propanoic acid

→ Could be identified through a neutralisation reaction using `text{Na}_2text{CO}_3`.

→ Effervescent reaction will result.

Propan-1-ol

→ Could be identified through an oxidation reaction using acidified dichromate.

→ The reaction would cause the solution to change from green to orange.

ENGINEERING, AE 2021 HSC 23d

Carbon fibre epoxy is an example of a composite material used in aircraft construction.

Describe the benefits of using this composite in aircraft construction in terms of both its manufacturing properties and its in-service properties.   (3 marks)

Show Answers Only

→ Carbon fibre epoxy matrices can be formed into more complex shapes and are stronger than alternative materials. This results in an improved aircraft performance.

→ Carbon fibre is often added to resin to produce a material with higher specific strength than its predecessor. This weight loss decreases fuel consumption or allows for an increased number of passengers.

→ In-service properties of composites include non-directional strength or highly directional strength behaviour depending on the composite’s nature and layup.

Show Worked Solution

→ Carbon fibre epoxy matrices can be formed into more complex shapes and are stronger than alternative materials. This results in an improved aircraft performance.

→ Carbon fibre is often added to resin to produce a material with higher specific strength than its predecessor. This weight loss decreases fuel consumption or allows for an increased number of passengers.

→ In-service properties of composites include non-directional strength or highly directional strength behaviour depending on the composite’s nature and layup.

ENGINEERING, AE 2021 HSC 23b

How has the work of aeronautical engineers helped to improve aircraft safety?   (3 marks)

Show Answers Only

Engineers are responsible for the following safety improvements:

→ Autopilot reduces risk of pilot error without removing pilot completely.

→ Changes in materials have created stronger, lighter, more durable aircraft.

→ Many built-in redundancies in case a system in the aircraft fails.

→ Safety and disaster mitigation systems in the event of an accident.
 

Successful answers could also include:

→ Changes in materials have created stronger, lighter, more durable aircraft.

→ Vigilance devices ensure pilots are fully awake and concentrated.

→ Knowledge of material failure may be used to mitigate mid-air incidents.

→ Engineers use knowledge of available materials to design new aircraft.

→ Ensure maintenance work is completed to ensure safe aircraft operation.

Show Worked Solution

Engineers are responsible for the following safety improvements:

→ Autopilot reduces risk of pilot error without removing pilot completely.

→ Changes in materials have created stronger, lighter, more durable aircraft.

→ Many built-in redundancies in case a system in the aircraft fails.

→ Safety and disaster mitigation systems in the event of an accident.
 

Successful answers could also include:

→ Changes in materials have created stronger, lighter, more durable aircraft.

→ Vigilance devices ensure pilots are fully awake and concentrated.

→ Knowledge of material failure may be used to mitigate mid-air incidents.

→ Engineers use knowledge of available materials to design new aircraft.

→ Ensure maintenance work is completed to ensure safe aircraft operation.

ENGINEERING, CS 2021 HSC 14 MC

The diagram shows a beam.

The beam's second moment of area is 76.96 × 106 mm4. It is subjected to a maximum bending moment of 250 kN m.

What is the maximum bending stress of the beam?

  1. 227.4 MPa
  2. 324.8 MPa
  3. 649.7 MPa
  4. 9745.3 MPa
Show Answers Only

`B`

Show Worked Solution

`M=250\ text{kNm}\ = 250 xx10^(-3)\ text{Nm}`

`y=(200)/(2)\ text{mm}\ =100 xx10^(-3)\ text{m}`

`I=76.96 xx10^(6)\ text{mm}^(4)=76.96 xx10^(-6)\ text{m}^(4)`

`S_text{bending}` `=(My)/I`  
  `=(250 xx10^(-3)xx100 xx10^(-3))/(76.96 xx10^(-6))`  
  `=324.8440…\ text{MPa}`  

 
`=>B`

ENGINEERING, TE 2021 HSC 11 MC

Which radio transmission method is most likely to have static interference?

  1. Amplitude modulation (AM)
  2. Frequency modulation (FM)
  3. Digital audio broadcast (DAB)
  4. Pulse width modulation (PWM)
Show Answers Only

`A`

Show Worked Solution

By Elimination:

→ Digital broadcasting will have very little interference (eliminate `C` and `D`)

→ The higher frequency of FM output results in less static than AM (eliminate `B`)

`=>A`

ENGINEERING, AE 2021 HSC 9 MC

A pitot tube supplies two pressure readings, total pressure and static pressure.

These pressure readings are then used to determine the

  1. sealed pressure.
  2. dynamic pressure.
  3. standard pressure.
  4. hydrostatic pressure.
Show Answers Only

`B`

Show Worked Solution

Dynamic Pressure = Total − Static

`=>B`

ENGINEERING, CS 2021 HSC 7 MC

The diagram shows a section of a pin jointed truss in equilibrium. The force in Member 1 is 200 kN in compression.
 

Which row of the table identifies the magnitude and nature of the forces in Member 2 and Member 3?
 

Show Answers Only

`A`

Show Worked Solution

→ The sum of horizontal forces and the sum of vertical forces must both be equal to 0.

`=>A`

ENGINEERING, PPT 2021 HSC 6 MC

A train has a total mass of 400 tonnes. It accelerates from rest to a speed of 25 m/s.

What is the kinetic energy of the train at the speed of 25 m/s?

  1. `125 × 10^(3)\ text{J}`
  2. `250 × 10^(3)\ text{J}`
  3. `125 × 10^(6)\ text{J}`
  4. `250 × 10^(6)\ text{J}`
Show Answers Only

`C`

Show Worked Solution

Using 1 tonne = 1000 kg:

`KE` `= 1/2 xx mxxv ^2`  
  `=0.5xx400\ 000 × 25^2`  
  `=125\ 000\ 000`  
  `=125 xx 10^6\ text{J}`  

 
`=>C`

BIOLOGY, M5 2020 HSC 20 MC

This chart illustrates three correlation patterns indicating the influence of genes and environment on different traits in individuals.
 

What does the data show about how genes and family environment affect the three traits?
 

Show Answers Only

`A`

Show Worked Solution

→ Trait `A` is similar in adoptive siblings that do not share much genetics.

→ Trait `B` is very similar in identical twins with identical genetics.

→ Trait `C` has little similarity between close relations or adoptive siblings.

`=>A`

BIOLOGY, M6 2020 HSC 17 MC

There are about 10 million single nucleotide polymorphisms (SNPs) found in the human genome.

Four SNPs are modelled in the diagram.
 

The SNPs modelled do not affect the phenotype of the individuals shown.

Which is the best explanation for this?

  1. Only one nucleotide is different at each SNP.
  2. The SNPs are part of DNA that is not expressed.
  3. AGA, CAA, TAT and CTC all code for the same amino acid.
  4. The SNPs are present on one strand of the DNA molecule only.
Show Answers Only

`B`

Show Worked Solution

→ Mutations that occur on non-coding DNA are not expressed as proteins.

→ Phenotype is therefore not affected.

`=>B`

BIOLOGY, M6 2020 HSC 13 MC

A type of genetic technology is shown in the diagram.
 


 

What type of cloning is modelled?

  1. Gene cloning because bacteria are used.
  2. Gene cloning because a human gene is being replicated.
  3. Whole organism cloning because identical offspring are produced.
  4. Whole organism cloning because the bacteria use asexual reproduction.
Show Answers Only

`B`

Show Worked Solution

→ A human gene is extracted and inserted into a bacterial plasmid.

→ This allows multiple copies of the gene to be produced via the process of gene cloning.

`=>B`

BIOLOGY, M8 2020 HSC 11 MC

The diagram shows a model of the human eye.
 

Which of the following correctly identifies a labelled part and its function?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \textit{Label} \quad \rule[-1ex]{0pt}{0pt}& \quad \textit{Name} \quad & \textit{Function} \\
\hline
\rule{0pt}{2.5ex}\text{I}\rule[-1ex]{0pt}{0pt}&\text{Cornea}&\text{Refract light}\\
\hline
\rule{0pt}{2.5ex}\text{I}\rule[-1ex]{0pt}{0pt}& \text{Retina}&\quad \text{Transmit light}\quad \\
\hline
\rule{0pt}{2.5ex}\text{II}\rule[-1ex]{0pt}{0pt}& \text{Retina} &\text{Focus light}\\
\hline
\rule{0pt}{2.5ex}\text{II}\rule[-1ex]{0pt}{0pt}& \text{Cornea} &\text{Absorb light}\\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

→ The cornea is the clear disc at the front of the eye, which helps protect the eye as well as refract incoming light rays.

\(\Rightarrow A\)

BIOLOGY, M8 2020 HSC 8 MC

Quarantine is ineffective as a measure to control non-infectious diseases because they

  1. cannot develop in isolation.
  2. depend on long-term exposure to a pathogen.
  3. may be inherited and affect the organism all their life.
  4. may only be treated by genetic engineering altering cells.
Show Answers Only

`C`

Show Worked Solution

→ Non-infectious diseases are often inherited (non-transmittable) diseases.

→ These affect an organism’s genome and are not transmitted or acquired by pathogens.

→ Quarantine would be an ineffective method of control.

`=>C`

BIOLOGY, M5 2020 HSC 5 MC

Which row of the table best describes DNA in both prokaryotic and eukaryotic cells?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex}\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex} \textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex} \textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex} \textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex} \textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex} \textit{Prokaryotic } \rule[-1ex]{0pt}{0pt}& \textit{Eukaryotic} \\
\hline \rule{0pt}{2.5ex} \text {Circular} \rule[-1ex]{0pt}{0pt}& \text {Circular} \\
\hline \rule{0pt}{2.5ex} \text {Circular} \rule[-1ex]{0pt}{0pt}& \text {Linear} \\
\hline \rule{0pt}{2.5ex} \text {Linear} \rule[-1ex]{0pt}{0pt}& \text {Circular} \\
\hline \rule{0pt}{2.5ex}  \text {Linear} \rule[-1ex]{0pt}{0pt}& \text {Linear} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution

→ DNA for prokaryotes is contained in a single circular chromosome, as well as in circular plasmids.

→ Eukaryotes have their main coding DNA within densely packed linear chromosomes.

\(\Rightarrow B\)

BIOLOGY, M5 2020 HSC 3 MC

The following four events occur during reproduction in a placental mammal.

  1. Fertilisation
  2. Implantation
  3. Ovulation
  4. Placental formation

In which order do these events occur?

  1. 2, 1, 3, 4
  2. 2, 4, 1, 3
  3. 3, 1, 2, 4
  4. 3, 2, 4, 1
Show Answers Only

`C`

Show Worked Solution

→ Ovulation (the production of the female gamete) must occur first, followed by fertilisation of the gamete.

→ Once it is fertilised, it implants itself in the uterine wall, followed by the formation of the placenta.

`=>C`

BIOLOGY, M5 2020 HSC 2 MC

Sexual reproduction in plants involves

  1. pollination caused by dispersal of seeds.
  2. cloning as it creates copies of the parent plant.
  3. mitosis leading to the formation of pollen grains.
  4. fertilisation as a result of fusion of male and female gametes.
Show Answers Only

`D`

Show Worked Solution

→ The key process in sexual reproduction is the fusion of male and female gametes.

`=>D`

BIOLOGY, M8 2021 HSC 32

The flow chart shows negative feedback by the hormones testosterone and inhibin in a human male.
 


 

Some athletes take anabolic steroids to increase their muscle mass and strength. These steroids may be testosterone or a synthetic modification of testosterone.

Explain the changes that would occur in the testes of a male athlete continuously taking anabolic steroids. Support your answer with reference to the flow chart.   (5 marks)

Show Answers Only

→ Naturally, as shown in the diagram above, males produce Releasing Hormone via the hypothalamus, which stimulates the anterior pituitary gland to produce hormones which further stimulate the testes to produce hormones inhibin and testosterone as well as sperm (via testosterone).

→ A male taking anabolic steroids similar to testosterone will see an initial increase in sperm and inhibin.

→ However, it will also impact the negative feedback on both the hypothalamus and anterior pituitary, reducing their release of respective stimulatory hormones, reducing sperm, testosterone and inhibin.

→ Although less inhibin would mean a smaller negative feedback on the anterior pituitary caused by this hormone, the ongoing dose of testosterone would ensure a large overall negative feedback effect. 

→ Over time, with repetitive and excessive use of anabolic steroids, males will produce less sperm, testosterone and inhibin, and the testes will lose their normal function.

Show Worked Solution

→ Naturally, as shown in the diagram above, males produce Releasing Hormone via the hypothalamus, which stimulates the anterior pituitary gland to produce hormones which further stimulate the testes to produce hormones inhibin and testosterone as well as sperm (via testosterone).

→ A male taking anabolic steroids similar to testosterone will see an initial increase in sperm and inhibin.

→ However, it will also impact the negative feedback on both the hypothalamus and anterior pituitary, reducing their release of respective stimulatory hormones, reducing sperm, testosterone and inhibin.

→ Although less inhibin would mean a smaller negative feedback on the anterior pituitary caused by this hormone, the ongoing dose of testosterone would ensure a large overall negative feedback effect. 

→ Over time, with repetitive and excessive use of anabolic steroids, males will produce less sperm, testosterone and inhibin, and the testes will lose their normal function.

♦ Mean mark 50%.

Calculus, EXT2 C1 2022 HSC 5 MC

If  `int_(a)^(x)f(t)dt=g(x)`, which of the following is a primitive of  `f(x)g(x)` ?

  1. `(1)/(2)[f(x)]^(2)`
  2. `(1)/(2)[f^(')(x)]^(2)`
  3. `(1)/(2)[g(x)]^(2)`
  4. `(1)/(2)[g^(')(x)]^(2)`
Show Answers Only

`C`

Show Worked Solution
`int_a^x f(t)\ dt` `=g(x)`  
`d/dx [int_a^xf(t)\ dt]` `=g^{′}(x)`  
`f(x)` `=g^{′}(x)`  
`f(x)*g(x)` `=g^{′}(x)g(x)`  
`int f(x)*g(x)\ dx` `=int g^{′}(x)g(x)\ dx`  
  `=1/2[g(x)]^2+C`  

 
`=>C`


COMMENT: Simple examples can illustrate the logic of line 3 of the worked solution.

PHYSICS, M5 2019 HSC 35

The apparatus shown is attached horizontally to the roof inside a stationary car. The plane of the protractor is perpendicular to the sides of the car.
 

The car was then driven at a constant speed `(v)`, on a horizontal surface, causing the string to swing to the right and remain at a constant angle `(theta)` measured with respect to the vertical.

Describe how the apparatus can be used to determine features of the car's motion. In your answer, derive an expression that relates a feature of the car's motion to the angle `theta`.   (4 marks)

Show Answers Only

→ The constant deflection of the mass to the right indicates that the car has a uniform acceleration to the left.

→ Since the car’s speed is constant, it must be travelling in uniform circular motion.

→ The radius of the car’s motion can be found in terms of `theta.`

As there is no vertical acceleration of the mass, the vertical component of its tension will balance its weight:

`T_y=Tcos theta=mg\ \ …\ (1)`

The centripetal force on the mass is given by the horizontal component of its tension:

`T_x=Tsin theta=ma=(mv^2)/r\ \ …\ (2)`

Substitute `T=(mg)/costheta`  from (1) into (2):

`T sin theta` `=(mv^2)/(r)`
`(mg xx sin theta)/(cos theta)` `=(mv^(2))/(r)`
`g xx tan theta` `=(v^(2))/(r)`
`r` `=(v^(2))/(g xx tan theta)`

 
Other expressions could include:

  • `v=sqrt(rg tan theta)`
Show Worked Solution

→ The constant deflection of the mass to the right indicates that the car has a uniform acceleration to the left.

→ Since the car’s speed is constant, it must be travelling in uniform circular motion.

→ The radius of the car’s motion can be found in terms of `theta.`

As there is no vertical acceleration of the mass, the vertical component of its tension will balance its weight:

`T_y=Tcos theta=mg\ \ …\ (1)`

The centripetal force on the mass is given by the horizontal component of its tension:

`T_x=Tsin theta=ma=(mv^2)/r\ \ …\ (2)`

Substitute `T=(mg)/costheta`  from (1) into (2):

`T sin theta` `=(mv^2)/(r)`
`(mg xx sin theta)/(cos theta)` `=(mv^(2))/(r)`
`g xx tan theta` `=(v^(2))/(r)`
`r` `=(v^(2))/(g xx tan theta)`

 
Other expressions could include:

  • `v=sqrt(rg tan theta)`

♦ Mean mark 52%.

PHYSICS, M6 2019 HSC 33

A proton and an alpha particle are fired into a uniform magnetic field with the same speed from opposite sides as shown. Their trajectories are initially perpendicular to the field.
 


  

Explain ONE similarity and ONE difference in their trajectories as they move in the magnetic field.   (4 marks)

Show Answers Only

Similarity:

→ Both particles experience a constant force, given by  `F=qvB`, perpendicular to their velocity and the magnetic field lines and will undergo circular motion.

Difference:

→ The centripetal force acting on both particles is given by the force they experience due to the magnetic field as follows:

`(mv^2)/(r)=qvB\ \ =>\ \ r=(mv)/(qB)`

→ The alpha particle has four times the mass and two times the charge of the proton

→ Therefore, the radius of its trajectory will be twice that of the protons.

Show Worked Solution

Similarity:

→ Both particles experience a constant force, given by  `F=qvB`, perpendicular to their velocity and the magnetic field lines and will undergo circular motion.

Difference:

→ The centripetal force acting on both particles is given by the force they experience due to the magnetic field as follows:

`(mv^2)/(r)=qvB\ \ =>\ \ r=(mv)/(qB)`

→ The alpha particle has four times the mass and two times the charge of the proton

→ Therefore, the radius of its trajectory will be twice that of the protons.

BIOLOGY, M6 2021 HSC 15 MC

An example of the mutagenic effect of ultraviolet radiation (UV) on DNA is shown in the diagram.
 

What is the mutagenic effect that is modelled?

  1. Thymine is duplicated.
  2. Bonds are formed between adjacent bases.
  3. Nucleotides form bonds in the backbone of DNA.
  4. Thymines on the two strands of DNA form bonds.
Show Answers Only

`B`

Show Worked Solution

→ The diagram shows bonds that have formed between adjacent thymine bases on a single strand of DNA.

`=>B`

Mean mark 56%.

BIOLOGY, M6 2021 HSC 11 MC

Many transgenic crops have been genetically engineered to have traits such as herbicide resistance. In at least four different crops the transgene has been found in nearby wild plant relatives of the cultivated crops.

What is the most likely reason for this observation?

  1. Crossing over in the wild plants
  2. Gene flow from the crops to the wild plants
  3. Genetic drift from the crops to the wild plants
  4. Mutations in the wild plants that match the transgenes
Show Answers Only

`B`

Show Worked Solution

→ Gene flow is the transfer of genetic material from one population to another.

→ Therefore the genetically engineered crops have most likely pollinated the nearby wild plant relatives.

`=>B`


Mean mark 57%.

Calculus, EXT2 C1 2022 HSC 4 MC

Of the following expressions, which one need NOT contain a term involving a logarithm in its anti-derivative?

  1. `(x+2)/(x^(2)+4x+5)`
  2. `(x+2)/(x^(2)-4x-5)`
  3. `(x-1)/(x^(3)-x^(2)+x-1)`
  4. `(x+1)/(x^(3)-x^(2)+x-1)`
Show Answers Only

`C`

Show Worked Solution

`text{Consider the denominator of}\ C:`

`x^(3)-x^(2)+x-1` `=x^2(x-1)+(x-1)`  
  `=(x^2+1)(x-1)`  

 
`(x-1)/(x^(3)-x^(2)+x-1)=(x-1)/((x^2+1)(x-1))=1/(x^2+1)`

`int 1/(x^2+1)\ dx=tan^(-1)(x)+c`

`=>C`

Proof, EXT2 P1 2022 HSC 3 MC

Let `A, B, P` be three points in three-dimensional space with `A \ne B`.

Consider the following statement.

If `P` is on the line `A B`, then there exists a real number `\lambda` such that  `vec{A P}=\lambda \vec{A B}`.

Which of the following is the contrapositive of this statement?

  1. If for all real numbers `\lambda, \vec{A P}=\lambda \vec{A B}`, then `P` is on the line `A B`.
  2. If for all real numbers `\lambda, \vec{A P} \ne \lambda \vec{A B}`, then `P` is not on the line `A B`.
  3. If there exists a real number `\lambda` such that  `\vec{A P}=\lambda \vec{A B}`, then `P` is on the line `A B`.
  4. If there exists a real number `\lambda` such that  `\vec{A P} \ne \lambda \vec{A B}`, then `P` is not on the line `A B`.
Show Answers Only

`B`

Show Worked Solution

`text{Statement:}`

`text{If}\ P\ text{is on}\ AB\ \ =>\ \ EE lambda\ \ text{such that}\ \ vec{A P}=\lambda \vec{A B}`

 
`text{Contrapositive statement:}`

`text{If}\ not \ EE lambda\ \ text{such that}\ \ vec{A P}=\lambda \vec{A B}\ \ =>\ \ P\ text{is not on}\ AB`

 
`text{In other words …}`

`text{If for all real numbers}\ lambda, \vec{A P} \ne \lambda \vec{A B}, text{then}\ P\ text{is not on the line}\ A B.`

`=>B`

Proof, EXT2 P1 2022 HSC 2 MC

The following proof aims to establish that  – 4 = 0

`text{Let}` `a=-4`    
`=>` `a^2 = 16 \ text{and} ` `\ 4a + 4 = -12` `text{Line 1}`
`=>` `a^2 + 4a + 4 =` `4` `text{Line 2}`
`=>` `(a + 2)^2 =` `2^2` `text{Line 3}`
`=>` `a + 2 =` `2` `text{Line 4}`
`=>` `a =` `0`  

 
At which line is the implication incorrect?

  1. Line 1
  2. Line 2
  3. Line 3
  4. Line 4
Show Answers Only

`D`

Show Worked Solution

`text{Consider Line 4:}`

`text{Given}\ \ (a + 2)^2=2^2\ \ =>\ \ a+2=+-2`

`=>D`

PHYSICS, M6 2020 HSC 33

A strong magnet of mass 0.04 kg falls 0.78 m under the action of gravity from position `X` above a hollow copper cylinder. It then travels a distance of 0.20 m through the cylinder from `Y` to `Z` before falling freely again.
 

The magnet takes 0.5 seconds to pass through the cylinder. The displacement-time graph of the magnet is shown.
 

Analyse the motion of the magnet by applying the law of conservation of energy.

Your analysis should refer to gravity and the copper cylinder, and include both qualitative and quantitative information.   (9 marks)

Show Answers Only

During the first 0.4 seconds:

→  The magnet is accelerating downwards at 9.8 m s–2 due to gravity.

→  The magnet’s gravitational potential energy is being converted into kinetic energy consistent with the law of conservation of energy.

→  Quantitatively:

`Delta E_(k)` `= Delta U`  
  `=mg Delta h`  
  `=0.04 xx9.8 xx0.78`  
  `=0.30576` J  

 

→  Hence 0.30576 J of the magnet’s gravitational potential energy is converted into kinetic energy as it falls under gravity.

 
As magnet reaches the copper cylinder:

→  Its downwards motion causes the formation of induced currents in the cylinder which produce a magnetic field that opposes the magnet’s motion (Lenz’s law).

→  This causes the magnet to decelerate to 0.4 m s–1 and lose kinetic energy. 

→  Finding the magnets speed before entering the copper cylinder:

`E_(k)` `=(1)/(2)mv^2`  
`0.30576` `=(1)/(2)xx 0.04xx v^2`  
`v` `=3.91`  m s–1  

 
→  Quantifying the kinetic energy loss:

`Delta E_(k)` `=(1)/(2)mv^2-(1)/(2)m u^2`  
  `=(1)/(2)xx 0.04xx 3.91^2-(1)/(2)xx 0.04xx 0.4^2`  
  `=0.30256` J  

 
→  Applying the law of conservation of energy shows that 0.30256 J of the magnet’s kinetic energy is being converted to heat energy within the cylinder as the magnet decelerates.

→  Finally, as the magnet passes through the copper cylinder, its gravitational potential energy decreases while its velocity remains constant.

→  Quantifying the decrease in gravitational potential energy:

`Delta U` `=mg Delta h`  
  `=0.04xx 9.8xx 0.2`  
  `=0.0784`  

 

→  Hence, 0.0784 J of the magnet’s gravitational potential energy is converted into heat energy in the cylinder, consistent with the law of conservation of energy.

Show Worked Solution

During the first 0.4 seconds:

→  The magnet is accelerating downwards at 9.8 m s–2 due to gravity.

→  The magnet’s gravitational potential energy is being converted into kinetic energy consistent with the law of conservation of energy.

→  Quantitatively:

`Delta E_(k)` `= Delta U`  
  `=mg Delta h`  
  `=0.04 xx9.8 xx0.78`  
  `=0.30576` J  

 

→  Hence 0.30576 J of the magnet’s gravitational potential energy is converted into kinetic energy as it falls under gravity.

 
As magnet reaches the copper cylinder:

→  Its downwards motion causes the formation of induced currents in the cylinder which produce a magnetic field that opposes the magnet’s motion (Lenz’s law).

→  This causes the magnet to decelerate to 0.4 m s–1 and lose kinetic energy. 

→  Finding the magnets speed before entering the copper cylinder:

`E_(k)` `=(1)/(2)mv^2`  
`0.30576` `=(1)/(2)xx 0.04xx v^2`  
`v` `=3.91` m s–1  

 
→  Quantifying the kinetic energy loss:

`Delta E_(k)` `=(1)/(2)mv^2-(1)/(2)m u^2`  
  `=(1)/(2)xx 0.04xx 3.91^2-(1)/(2)xx 0.04xx 0.4^2`  
  `=0.30256` J  

 
→  Applying the law of conservation of energy shows that 0.30256 J of the magnet’s kinetic energy is being converted to heat energy within the cylinder as the magnet decelerates.

→  Finally, as the magnet passes through the copper cylinder, its gravitational potential energy decreases while its velocity remains constant.

→  Quantifying the decrease in gravitational potential energy:

`Delta U` `=mg Delta h`  
  `=0.04xx 9.8xx 0.2`  
  `=0.0784`  

 

→  Hence, 0.0784 J of the magnet’s gravitational potential energy is converted into heat energy in the cylinder, consistent with the law of conservation of energy.

♦ Mean mark 51%.

PHYSICS M5 2022 HSC 35

A capsule travels around the International Space Station (ISS) in a circular path of radius 200 m as shown.
 


 

Analyse this system to test the hypothesis below.  (5 marks)

The uniform circular motion of the capsule around the ISS can be accounted for in terms of the gravitational force between the capsule and the ISS.

Show Answers Only

Find the gravitational force between the capsule and the ISS:

`F` `=(GMm)/(r^2)`  
  `=(6.67 xx10^-11 xx4.2 xx10^5 xx 1.2 xx10^4)/(200^2)`  
  `=8.4 xx10^(-6)  text{N}`  

 
Find the centripetal force required to keep the capsule in its orbit:

`F_(c)` `=(mv^2)/(r)`  
  `=(1.2 xx10^4 xx 0.233^2)/(200)`  
  `=3.26  text{N}`  

 

→ The gravitational force is not sufficient to provide the necessary centripetal force to keep the capsule in orbit around the ISS.

→ The hypothesis is invalid.

Show Worked Solution

Find the gravitational force between the capsule and the ISS:

`F` `=(GMm)/(r^2)`  
  `=(6.67 xx10^-11 xx4.2 xx10^5 xx 1.2 xx10^4)/(200^2)`  
  `=8.4 xx10^(-6)  text{N}`  

 
Find the centripetal force required to keep the capsule in its orbit:

`F_(c)` `=(mv^2)/(r)`  
  `=(1.2 xx10^4 xx 0.233^2)/(200)`  
  `=3.26  text{N}`  

 

→ The gravitational force is not sufficient to provide the necessary centripetal force to keep the capsule in orbit around the ISS.

→ The hypothesis is invalid.

PHYSICS M6 2022 HSC 34

Three charged particles, `X, Y` and `Z`, travelling along straight, parallel trajectories at the same speed, enter a region in which there is a uniform magnetic field which causes them to follow the paths shown. Assume that the particles do not exert any significant force on each other.
 


 

Explain the different paths that the particles follow through the magnetic field.   (7 marks)

Show Answers Only

→ The charged particles each experience a force equal to  `F=qvB`  as they travel perpendicular to the magnetic field lines.

→ This force acts perpendicular to the direction of their velocity, so the charged particles undergo uniform circular motion with the force due to the magnetic field acting as a centripetal force.

→ The radii of their paths can be described by the equation:

`F_(c)` `=F_(b)`  
`(mv^2)/(r)` `=qvB`  
`r` `=(mv)/(qB)`  

 
→ As the strength of the magnetic field `(B)` and the speed `(v)` of the particles is the same, their trajectory is only affected by their mass to charge ratio `((m)/(q))`  which affects their radius of motion and their sign, which determines the direction which they deflect.

→ Both `X` and `Y` initially curve upwards, using the right hand palm rule, these must both be positive charges.

→  `Y` has a greater radius than `X`, so `Y`’s charge to mass ratio is greater than `X`’s.

→ `Z` initially curves downwards, using the right hand palm rule, it must have a negative charge.

→ `Z` has the same radius of motion as `X`, so the charge to mass ratios of `Z` and `X` are the same.

→ `Z` has a smaller radius than `Y`, so `Z`’s charge to mass ratio is less than that of `Y`.

Show Worked Solution

→ The charged particles each experience a force equal to  `F=qvB`  as they travel perpendicular to the magnetic field lines.

→ This force acts perpendicular to the direction of their velocity, so the charged particles undergo uniform circular motion with the force due to the magnetic field acting as a centripetal force.

→ The radii of their paths can be described by the equation:

`F_(c)` `=F_(b)`  
`(mv^2)/(r)` `=qvB`  
`r` `=(mv)/(qB)`  

 
→ As the strength of the magnetic field `(B)` and the speed `(v)` of the particles is the same, their trajectory is only affected by their mass to charge ratio `((m)/(q))`  which affects their radius of motion and their sign, which determines the direction which they deflect.

→ Both `X` and `Y` initially curve upwards, using the right hand palm rule, these must both be positive charges.

→  `Y` has a greater radius than `X`, so `Y`’s charge to mass ratio is greater than `X`’s.

→ `Z` initially curves downwards, using the right hand palm rule, it must have a negative charge.

→ `Z` has the same radius of motion as `X`, so the charge to mass ratios of `Z` and `X` are the same.

→ `Z` has a smaller radius than `Y`, so `Z`’s charge to mass ratio is less than that of `Y`.


Mean mark 60%.

PHYSICS M5 2022 HSC 33

In a hammer throw sport event, a 7.0 kg projectile rotates in a circle of radius 1.6 m, with a period of 0.50 s. It is released at point `P`, which is 1.2 m above the ground, where its velocity is at 45° to the horizontal.
 


 

  1. Show that the vertical component of the projectile's velocity at `P` is `14.2 \ text{m s}^(-1)`.  (2 marks)
  1. Calculate the horizontal range of the projectile from point `P`.  (4 marks)
Show Answers Only

a.   Proof (See Worked Solution)

b.   42.3 m

Show Worked Solution
a.    `v` `=(2pir)/(T)`
    `=(2 xxpi xx1.6)/(0.5)`
    `=20.106  text{ms}^(-1)`
     `v_(y)` `=20.106 xx sin 45^@`
    `=14.2\ text{ms}^(-1)`

 

b.   Time of flight `=>`  Find  `t`  when  `s_(y)=-1.2`

`s_(y)` `=u_(y)t+(1)/(2)a_(y)t^2`
`-1.2` `=14.2t-(1)/(2) xx9.8t^2`
`0` `=4.9t^2-14.2t-1.2`

 

`t` `=(14.2+-sqrt(14.2^2+4 xx4.9 xx1.2))/(2 xx4.9)`
  `=2.98  text{s}\ \ (t>0)`

 
Since launch angle = 45°  `=>\ \ u_(x)=u_(y)`

`:. s_(x)` `=u_(x)t`
  `=14.2 xx2.98`
  `=42.3  text{m}`

PHYSICS, M6 2022 HSC 32

One type of stationary exercise bike uses a pair of strong, movable magnets placed on opposite sides of a thick, aluminium flywheel to provide a torque to make it harder to pedal.
 


 

  1. Explain the principle by which these magnets make it harder to pedal.  (3 marks)
  1. The bike rider wants to increase the opposing torque on the flywheel. Justify an adjustment that could be made to the magnets to achieve this.  (3 marks)
Show Answers Only

a.    Magnetic breaking is a consequence of Lenz’s Law.

→ The changing magnetic flux through the wheel, as a result of its rotation, causes eddy currents to be induced (Faraday’s Law).

→ According to Lenz’s Law, these eddy currents produce a force which opposes the rotation of the wheel, making it more difficult to pedal.
 

b.   Adjustment to increase opposing torque:

→ Moving the magnets closer to the outer edge of the flywheel will increase the opposing torque on it. As the linear speed of the wheel is greater near the edge, the rate of change of flux passing through it will increase.

→ This increases the magnitude of induced emf as  `epsilon=-N(Delta theta)/(Delta t).`

→ Consequently, the opposing force and torque will increase.

→ Moving the magnets closer to the edge of the flywheel also increases the distance between the point of application of the opposing force and the axis of rotation of the wheel. As  `tau=rF`  this further increases opposing torque.

Show Worked Solution

a.    Magnetic breaking is a consequence of Lenz’s Law.

→ The changing magnetic flux through the wheel, as a result of its rotation, causes eddy currents to be induced (Faraday’s Law).

→ According to Lenz’s Law, these eddy currents produce a force which opposes the rotation of the wheel, making it more difficult to pedal.
 

b.   Adjustment to increase opposing torque:

→ Moving the magnets closer to the outer edge of the flywheel will increase the opposing torque on it. As the linear speed of the wheel is greater near the edge, the rate of change of flux passing through it will increase.

→ This increases the magnitude of induced emf as  `epsilon=-N(Delta theta)/(Delta t).`

→ Consequently, the opposing force and torque will increase.

→ Moving the magnets closer to the edge of the flywheel also increases the distance between the point of application of the opposing force and the axis of rotation of the wheel. As  `tau=rF`  this further increases opposing torque.

PHYSICS M8 2022 HSC 31

Following the Geiger-Marsden experiment, Rutherford proposed a model of the atom.
 


 

Bohr modified this model to explain the spectrum of hydrogen observed in experiments.
 


 

The Bohr-Rutherford model of the atom consists of electrons in energy levels around a positive nucleus.

How do features of this model account for all the experimental evidence above? Support your answer with a sample calculation and a diagram, and refer to energy, forces and photons.  (9 marks)

Show Answers Only

The Geiger-Marsden experiment, which involved firing alpha particles at a thin sheet of gold foil produced results which can be explained by the Bohr-Rutherford model:

→ The majority of fired alpha particles passed through the gold foil undeflected. Rutherford concluded from this that the atom had a small, central nucleus.

→ Some alpha particles were deflected and some of these were deflected at very large angles. Rutherford concluded from this that the nucleus was dense and positively charged exerting a repulsive electromagnetic force on the fired alpha particles.

→ The model accounts for Rutherford’s conclusions, placing electrons in orbits around a small positive nucleus.
 

Rutherford’s model alone could not explain the emission spectra of elements such as hydrogen. Bohr’s contribution to the Bohr-Rutherford model amended this:
 


 

→ Bohr proposed that electrons orbited the atomic nucleus in quantised orbits at fixed energies. He proposed that electrons could move from a higher energy orbit (eg. n=1) to a lower energy orbit (n=3) by emitting a photon with energy  `E=hf`  equal to the energy difference between the two orbits.

→ Additionally, he proposed that electrons could move from a lower energy orbit to a higher energy orbit by absorbing a photon with energy  `E=hf`  equal to the energy difference between the two orbits. 

→ This is able to account for the given emission spectra of hydrogen, where emission lines correspond to electron transitions from higher energy orbits to the second energy orbit which produce photons within the spectrum of visible light.
 

Using Rydberg’s equation it is possible to predict the emission lines of hydrogen, using an electron moving from the sixth to the second Bohr energy orbit as an example:
 

`(1)/(lambda)` `=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`  
  `=(1.097 xx10^7)((1)/(2^(2))-(1)/(6^(2)))`  
  `=(2 xx1.097 xx10^7)/(9)`  
  `=2.438 xx10^6`  
`lambda` `=410  text{nm}`  

 
→ This value corresponds to the leftmost line on the given spectrum, reflecting how Bohr’s model can account for the emission spectra of hydrogen.

 

Show Worked Solution

The Geiger-Marsden experiment, which involved firing alpha particles at a thin sheet of gold foil produced results which can be explained by the Bohr-Rutherford model:

→ The majority of fired alpha particles passed through the gold foil undeflected. Rutherford concluded from this that the atom had a small, central nucleus.

→ Some alpha particles were deflected and some of these were deflected at very large angles. Rutherford concluded from this that the nucleus was dense and positively charged exerting a repulsive electromagnetic force on the fired alpha particles.

→ The model accounts for Rutherford’s conclusions, placing electrons in orbits around a small positive nucleus.
 

Rutherford’s model alone could not explain the emission spectra of elements such as hydrogen. Bohr’s contribution to the Bohr-Rutherford model amended this:
 


 

→ Bohr proposed that electrons orbited the atomic nucleus in quantised orbits at fixed energies. He proposed that electrons could move from a higher energy orbit (eg. n=1) to a lower energy orbit (n=3) by emitting a photon with energy  `E=hf`  equal to the energy difference between the two orbits.

→ Additionally, he proposed that electrons could move from a lower energy orbit to a higher energy orbit by absorbing a photon with energy  `E=hf`  equal to the energy difference between the two orbits. 

→ This is able to account for the given emission spectra of hydrogen, where emission lines correspond to electron transitions from higher energy orbits to the second energy orbit which produce photons within the spectrum of visible light.
 

Using Rydberg’s equation it is possible to predict the emission lines of hydrogen, using an electron moving from the sixth to the second Bohr energy orbit as an example:
 

`(1)/(lambda)` `=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`  
  `=(1.097 xx10^7)((1)/(2^(2))-(1)/(6^(2)))`  
  `=(2 xx1.097 xx10^7)/(9)`  
  `=2.438 xx10^6`  
`lambda` `=410  text{nm}`  

 
→ This value corresponds to the leftmost line on the given spectrum, reflecting how Bohr’s model can account for the emission spectra of hydrogen.


♦ Mean mark 51%.

PHYSICS M7 2022 HSC 30

In a thought experiment, light travels from `X` to a mirror `Y` and back to `X` on a moving train carriage. The path of the light relative to an observer on the train is shown.
 


 

Relative to an observer outside the train, the path of the light is shown below, at three consecutive times as the train carriage moves along the track.
 

  1. Describe qualitatively how the constancy of the speed of light and the thought experiment above led Einstein to predict time dilation.  (3 marks)

  1. The train is travelling with a velocity  `v=0.96 c`. To the observer inside the train, the return journey for the light between `X` and `Y` takes 15 nanoseconds.

  2. How long would this return journey take according to the observer outside the train?  (3 marks)

Show Answers Only

a.  Consider each observer:

→ The observer on the train sees light travel a distance from `X` to `Y` and back to `X` again.

→ The observer outside the train sees light travel a longer path due to the horizontal motion of the train.

→ As the speed of light is constant for both observers, the observer outside the train must observe the light pulse to take a longer time, `t=(text{distance})/(c).`

→ This provided the basis for Einstein’s predictions of time dilation.
 

b.  53.6  ns

Show Worked Solution

a.  Consider each observer:

→ The observer on the train sees light travel a distance from `X` to `Y` and back to `X` again.

→ The observer outside the train sees light travel a longer path due to the horizontal motion of the train.

→ As the speed of light is constant for both observers, the observer outside the train must observe the light pulse to take a longer time, `t=(text{distance})/(c).`

→ This provided the basis for Einstein’s predictions of time dilation.


♦ Mean mark (a) 48%.
b.  `t`

 
 
`=(t_(0))/(sqrt(1-(v^(2))/(c^(2))))`
    `=(15)/(sqrt(1-((0.96c)^(2))/(c^(2))))`
    `=53.6  text{ns}`

PHYSICS M7 2022 HSC 27

A laser producing red light of wavelength 655 nm is directed onto double slits separated by a distance, `d=5.0 xx 10^{-5} \ text{m}`. A screen is placed behind the double slits.
 


 

  1. Newton proposed a model of light. Use a labelled sketch to show the pattern on the screen that would be expected from Newton's proposed model.   (2 marks)
     

     
  2. When the laser light is turned on, a series of vertical bright lines are seen on the screen.
     
 
  1. Calculate the angle, `\theta`, between the centre line and the bright line at `A`.   (3 marks)
  1. The laser is replaced with one producing green light of wavelength 520 nm.
  2. Explain the difference in the pattern that would be produced.   (2 marks)
Show Answers Only

a.   
       

 
b.   `theta=1.50^@`

c.   Consider  `d sin theta = m lambda :`

→ `sin theta prop lambda`

→ Using light with a shorter wavelength decreases the angular separation of bright fringes.

→ The bright lines will appear closer together.

Show Worked Solution
a.   
     

Mean mark part (a) 53%.

b.   Using  `d  sin  theta=mlambda:`

`5.0 xx10^(-5)  sin  theta` `=2 xx6.55xx10^(-7)`
   `sin  theta` `=(2 xx6.55xx10^(-7))/(5.0 xx10^(-5))`
   `theta` `=1.50^@`

 

c.   Consider  `d sin theta = m lambda :`

→ `sin theta prop lambda`

→ Using light with a shorter wavelength decreases the angular separation of bright fringes.

→ The bright lines will appear closer together.

PHYSICS, M7 2018 HSC 2 MC

Which graph is consistent with predictions resulting from Planck's hypothesis regarding radiation from hot objects?
 

 

Show Answers Only

`B`

Show Worked Solution

→ Planck’s hypothesis is consistent with Wein’s Law,  `lambda_(max)=(b)/(T)`, or  `lambda_(max) prop (1)/(T)`.

→ So, hotter objects have a greater peak wavelength.

`=>B`

PHYSICS, M7 2016 HSC 11 MC

What is the wavelength, in metres, of a photon with an energy of 3.5 eV?

  1. `1.2 × 10^(-6)`
  2. `3.5 × 10^(-7)`
  3. `1.18 × 10^(-15)`
  4. `5.67 × 10^(-26)`
Show Answers Only

`B`

Show Worked Solution
`E` `=3.5 xx1.602 xx10^(-19)`  
  `=5.607 xx10^(-19)  text{J}`  

 
`E=hf=(hc)/(lambda)\ \ => \ \ lambda=(hc)/(E)`

`:.lambda` `=(6.626 xx10^(-34)xx3xx10^(8))/(5.607 xx10^(-19))`  
  `=3.5 xx10^(-7)  text{m}`  

 
`=>B`

PHYSICS, M7 2016 HSC 13 MC

When light of a specific frequency strikes a metal surface, photoelectrons are emitted.

If the light intensity is increased but the frequency remains the same, which row of the table is correct?

\begin{align*}
\begin{array}{l}
\rule{0pt}{1.5ex}\text{} & \text{} \\
\text{}\rule[-0.5ex]{0pt}{0pt}& \text{} \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{1.5ex}\textit{Number of photoelectrons} & \textit{Maximum kinetic energy} \\
\textit{emitted}\rule[-0.5ex]{0pt}{0pt}& \textit{of the photoelectrons} \\
\hline
\rule{0pt}{2.5ex}\text{Remains the same}\rule[-1ex]{0pt}{0pt}&\text{Remains the same}\\
\hline
\rule{0pt}{2.5ex}\text{Remains the same}\rule[-1ex]{0pt}{0pt}& \text{Increases}\\
\hline
\rule{0pt}{2.5ex}\text{Increases}\rule[-1ex]{0pt}{0pt}& \text{Remains the same} \\
\hline
\rule{0pt}{2.5ex}\text{Increases}\rule[-1ex]{0pt}{0pt}& \text{Increases} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution

→ An increase in light intensity increases the number of photons striking the metal surface.

→ This increases the number of photoelectrons emitted.

→ Since no change in frequency, the energy of photons striking metal surface remains constant.

→ Maximum kinetic energy of emitted photoelectrons remains the same.

\(\Rightarrow C\)

PHYSICS, M7 2018 HSC 17 MC

The graph shows the maximum kinetic energy of electrons ejected from different metals as a function of the frequency of the incident light.
 

What can be deduced from this graph?

  1. The maximum kinetic energy of ejected electrons is proportional to the number of photons incident on the metal surface.
  2. More photons are required to cause an electron to be ejected from zinc than from potassium.
  3. Any photon that can eject an electron from the surface of zinc must also be able to cause an electron to be ejected from potassium.
  4. For any given frequency that causes electrons to be ejected from all three metals, the number of electrons ejected is always greatest for potassium.
Show Answers Only

`C`

Show Worked Solution

→ The graph shows that zinc has a greater work function than potassium.

→ So, any photon with sufficient energy to eject a photon from the surface of zinc will also have sufficient energy to overcome the work function of potassium and eject a photon from its surface.

`=>C`

PHYSICS, M7 2014 HSC 26

  1. Calculate the energy of a photon of wavelength 415 nm.  (2 marks)
  2. An experiment was conducted using a photoelectric cell as shown in the diagram.

The graph plots the maximum kinetic energy of the emitted photoelectrons against radiation frequency for the aluminium surface.

The experiment is planned to be repeated using a voltage of 0.0 V.

Draw a line on the graph to show the predicted results of the planned experiment, and determine the radiation frequency which would produce photoelectrons with a maximum kinetic energy of 1.2 eV using a voltage of 0.0 V.  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution
a.   `c` `= f lambda`
    `= hf`
    `= (hc)/lambda`
    `= (6.626 xx 10^(-34) Js xx 3 xx 10^8 ms^(-1))/(415 xx 10^(-9)m)`
    `= 4.79 xx 10^(-19) J`

 

b.  

`text(To find intercept)`

`4.1 text(V)` `= 4.1 xx 1.602 xx 10^(-19)\ text(J)` `text(of energy required to be)`
`text(supplied by the photon.)`
  `= 6.56 xx 10^(-19)\ text(J)`  
`hf` `= 6.56 xx 10^(-19)\ text(J)`  
`f` `= (6.56 xx 10^(-19))/(6.626 xx 10^(-34))`  
  `= 9.9 xx 10^14`  

`text(Gradient = same as A1)`

`text{From graph maximum KE(eV)}` `= 1.2\ text(eV)`
`text(Frequency)` `= 12.8\ text(Hz)`
  `= 12.8 xx 10^14\ text(Hz)`

PHYSICS M7 2022 HSC 26

Light of frequency  `7.5 xx10^(14) \ text{Hz}`  is incident on a calcium metal sheet which has a work function of  `2.9 \ text{eV}`. Photoelectrons are emitted.

The metal is in a uniform electric field of `5.2 \ text{NC}^{-1}`, perpendicular to the surface of the metal, as shown.
 


  

  1. Show that the maximum kinetic energy of an emitted photoelectron is  `3.2 xx10^(-20) \ text{J}`.  (3 marks)
  1. Calculate the maximum distance, `d`, an emitted photoelectron can travel from the surface of the metal.  (3 marks)
Show Answers Only
  1. Proof (See Worked Solution)
  2. `0.038  text{m.}`
Show Worked Solution
a.    `K_(max)` `=hf-Phi`
    `=6.626 xx10^(-34)xx7.5 xx10^(14)-2.9xx 1.602 xx10^(-19)`
    `=3.2 xx10^(-20)  text{J}`

 

b.   Maximum Distance:

→ The electric field will do `3.2 xx10^(-20)\ text{J}` of work to stop the electron.

`W` `=qEd`     
`3.2 xx10^(-20)` `=1.602 xx10^(-19) xx 5.2 xx d_max`  
`:.d_max` `=(3.2 xx10^(-20))/(1.602 xx10^(-19)xx 5.2)`  
  `=0.038  text{m.}`  

♦ Mean mark (b) 51%.