SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

PHYSICS, M8 2015 HSC 34aii

The mass of a polonium-218 nucleus is 218.00897 u, the mass of a lead-214 nucleus is 213.99981 u, and the mass of an alpha particle is 4.00260 u.

Calculate the energy released by this alpha decay.   (3 marks)

Show Answers Only

`9.8 xx10^(-13)\ text{J}`

Show Worked Solution

\(\text{Mass}_{i} = 218.00897\ u \)

\(\text{Mass}_{f} = 213.99981 + 4.00260 = 218.00241\ u \)

\(\text{Difference} = 218.00897-218.00241 = 0.00656\ u \)

\(\text{Difference (kg)} = 0.00656 \times 1.661 \times 10^{-27} = 1.08962 \times 10^{-29}\ \text{kg} \)
 

`E` `=mc^2`  
  `=1.08962xx10^(-29)xx(3 xx10^8)^2`  
  `=9.8 xx10^(-13)\ text{J}`  

PHYSICS, M8 2017 HSC 34bii

The diagram shows some components of a nuclear reactor.
 

Explain how the labelled components work together to produce a controlled nuclear reaction.   (3 marks)

Show Answers Only

→ Fuel elements are composed of a fissile material that release energy and high-velocity neutrons when atoms are split.

→ These high speed neutrons are slowed down by the moderator material. 

→ In this way, the neutrons are able to be captured by more fissile atoms, which in turn split.

→ The control rods play an important role in the process by absorbing neutrons. This limits the number of atoms undergoing fission and hence they can be used to control the release of energy.

Show Worked Solution

→ Fuel elements are composed of a fissile material that release energy and high-velocity neutrons when atoms are split.

→ These high speed neutrons are slowed down by the moderator material. 

→ In this way, the neutrons are able to be captured by more fissile atoms, which in turn split.

→ The control rods play an important role in the process by absorbing neutrons. This limits the number of atoms undergoing fission and hence they can be used to control the release of energy.

PHYSICS, M8 2018 HSC 34bii

The following is a nuclear reaction that produces a neutron.

`\ _(2)^(4)alpha + \ _(4)^(9)text{Be}rarr \ _(6)^(12)text{C} + \ _(0)^(1)text{n}`

The table shows the masses of the particles in the reaction.

         

Using the data from the table, calculate the energy released in this reaction. State your answer in joules.   (3 marks)

Show Answers Only

`7.0 xx10^(-13)\ text{J}`

Show Worked Solution

\(\text{Mass}_{i} = 4.0012 + 9.0122 = 13.0134\ u \)

\(\text{Mass}_{f} = 12.000 + 1.0087 = 13.0087\ u \)

\(\text{Difference} = 13.0087-13.0134 = –0.0047\ u \)

\(\text{Difference (kg)} = 0.0047 \times 1.661 \times 10^{-27} = 7.8067 \times 10^{-30}\ \text{kg} \)
 

`E` `=mc^2`  
  `=7.8067xx10^(-30)xx(3 xx10^8)^2`  
  `=7.0 xx10^(-13)\ text{J}`  

PHYSICS, M8 2018 HSC 34bi

The diagram shows apparatus used to investigate subatomic particles.
 

     

How did Chadwick use a law of physics to identify a property of `X`?   (2 marks)

Show Answers Only

→ Chadwick measured the momentum of both `X` and the ejected protons.

→ Using the law of conservation of momentum he was able to calculate the mass of `X` which was similar to the mass of a proton.

Show Worked Solution

→ Chadwick measured the momentum of both `X` and the ejected protons.

→ Using the law of conservation of momentum he was able to calculate the mass of `X` which was similar to the mass of a proton.

PHYSICS, M8 2019 HSC 36

A radon-198 atom, initially at rest, undergoes alpha decay. The masses of the atoms involved are shown in atomic mass units `(u)`.  
 

The kinetic energy of the polonium atom produced is  `2.55 × 10^(-14)` J.

By considering mass defect, calculate the kinetic energy of the alpha particle, and explain why it is significantly greater than that of the polonium atom.   (7 marks)

Show Answers Only
`Delta m` `=197.999-(193.988+4.000260)`  
  `=0.0084` u  
  `=0.0084 xx1.661 xx10^(-27)` (Converting to kg)
  `=1.395 xx10^(-29)` kg  

 

`E` `=mc^2`  
  `=1.395 xx10^(-29)xx(3xx10^(8))^(2)`  
  `=1.256 xx10^(-12)` J  

 
Applying the law of conservation of energy:

`KE_(text{alpha})` `=E_(text{total})-KE_(polonium)`  
  `=1.256 xx10^(-12)-2.44 xx10^(-14)`  
  `=1.23 xx10^(-12)` J  

 
→ As the radon atom is initially at rest, the initial momentum of this reaction system is zero. So, by applying the principle of conservation of momentum, the decay products must move away from each other with equal and opposite momenta.

→ The alpha particle has a significantly lower mass compared to the polonium atom and therefore has a significantly higher velocity.

→ `KE=(1)/(2)mv^(2)=(1)/(2)mv xx v`  and the momenta, `mv`  are equal for both the alpha particle and polonium atom. The alpha particle therefore has a significantly greater kinetic energy due to its greater velocity.

Show Worked Solution
`Delta m` `=197.999-(193.988+4.000260)`  
  `=0.0084` u   
  `=0.0084 xx1.661 xx10^(-27)` (Converting to kg)
  `=1.395 xx10^(-29)` kg  

 

`E` `=mc^2`  
  `=1.395 xx10^(-29)xx(3xx10^(8))^(2)`  
  `=1.256 xx10^(-12)` J  

 
Applying the law of conservation of energy:

`KE_(text{alpha})` `=E_(text{total})-KE_(polonium)`  
  `=1.256 xx10^(-12)-2.44 xx10^(-14)`  
  `=1.23 xx10^(-12)` J  

 
→ As the radon atom is initially at rest, the initial momentum of this reaction system is zero. So, by applying the principle of conservation of momentum, the decay products must move away from each other with equal and opposite momenta.

→ The alpha particle has a significantly lower mass compared to the polonium atom and therefore has a significantly higher velocity.

→ `KE=(1)/(2)mv^(2)=(1)/(2)mv xx v`  and the momenta, `mv`  are equal for both the alpha particle and polonium atom. The alpha particle therefore has a significantly greater kinetic energy due to its greater velocity.


♦♦♦ Mean mark 30%.

PHYSICS, M8 2019 HSC 34

Use the following information to answer this question.
 

Describe both the production and radiation of energy by the sun. In your answer, include a quantitative analysis of both the power output and the surface temperature of the sun.   (9 marks)

Show Answers Only

Production of the sun’s energy is predominantly done through the proton-proton chain. Four protons react in the sun’s core to produce one helium atom. The mass of one helium atom is less than the mass of four protons so mass is converted into energy through Einstein’s mass-energy equivalence, `E=mc^2.`

The sun acts as a black body and so radiates its energy in the form of black body radiation. It radiates energy as light with a variety of wavelengths, as shown in the black body curve above. It peaks at a specific wavelength which can be used to calculate its temperature:

`lambda_(max)=5.0xx10^-7` m  (see graph above)

`lambda_(max)` `=(b)/(T)`  
`T` `=(2.898xx10^-3)/(5.0xx10^-7)`  
  `=5796` K  
  `=5800` K  

 
Energy radiated from the sun spreads out over a sphere as it travels away from the sun. The intensity of radiated energy decreases at distances further from the sun, consistent with the inverse square law.

Using the intensity of the sun’s radiation at earth, its power output can be calculated:

`P` `=IA`  
  `=Ixx4pi r^2`  
  `=1360 xx4xx3.142 xx(1.5 xx10^(11))^(2)`  
  `=3.85 xx10^(26)` W  
Show Worked Solution

Production of the sun’s energy is predominantly done through the proton-proton chain. Four protons react in the sun’s core to produce one helium atom. The mass of one helium atom is less than the mass of four protons, so mass is converted into energy through Einstein’s mass-energy equivalence, `E=mc^2.`

The sun acts as a black body and so radiates its energy in the form of black body radiation. It radiates energy as light with a variety of wavelengths, as shown in the black body curve above. It peaks at a specific wavelength which can be used to calculate its temperature:

`lambda_(max)=5.0xx10^-7` m  (see graph above)

`lambda_(max)` `=(b)/(T)`  
`T` `=(2.898xx10^-3)/(5.0xx10^-7)`  
  `=5796` K  
  `=5800` K  

 
Energy radiated from the sun spreads out over a sphere as it travels away from the sun. The intensity of radiated energy decreases at distances further from the sun, consistent with the inverse square law.

Using the intensity of the sun’s radiation at earth, its power output can be calculated:

`P` `=IA`  
  `=Ixx4pi r^2`  
  `=1360 xx4xx3.142 xx(1.5 xx10^(11))^(2)`  
  `=3.85 xx10^(26)` W    

♦♦♦ Mean mark 38%.

PHYSICS M8 2022 HSC 28

Two steps in the CNO cycle of nuclear fusion are shown.
 

Step `X` releases 1.20 MeV.

The masses in Step `Y` are shown in the table.
 

Propose a reason why Step `Y` releases more energy than Step `X`. Support your answer with calculations.  (3 marks)

Show Answers Only

→ The energy released by a nuclear reaction is proportional to the mass defect.

→ Step `Y` will release more energy if it has a greater mass defect than step `X`.
 

Mass defect (Step `Y`) `=m_(reactants)-m_(pr oducts)`  
  `=1.007 + 13.003-14.003`  
  `=0.007u`  

 

`E` `=0.007 xx931.5`  
  `=6.5  text{MeV}`  

  
Which is greater than the energy released by step `X`.

Show Worked Solution

The energy released by a nuclear reaction is proportional to the mass defect. Step `Y` will release more energy if it has a greater mass defect than step `X`.
 

Mass defect (Step `Y`) `=m_(reactants)-m_(pr oducts)`  
  `=1.007 + 13.003-14.003`  
  `=0.007u`  

 

`E` `=0.007 xx931.5`  
  `=6.5  text{MeV}`  

  
Which is greater than the energy released by step `X`.


Mean mark 51%.

PHYSICS M8 2022 HSC 16 MC

The binding energy of helium-4 (He-4) is 28.3 MeV and the binding energy of beryllium-6 (Be-6) is 26.9 MeV.

Which of the following rows in the table is correct?
 

Show Answers Only

`A`

Show Worked Solution

The binding energy of a nucleus is the energy required to separate it into individual particles.

→ He-4 requires more energy to separate into individual protons and neutrons.

→ He-4 has 4 nucleons while Be-6 has 6 nucleons.

→ He-4 is less massive.

`=>A`

PHYSICS, M8 2019 HSC 19 MC

Consider the following nuclear reaction.

W  `+`  X  `→`  Y  `+`  Z

Information about W, X and Y is given in the table.
 

Which of the following is a correct statement about energy in this reaction?

  1. The reaction gives out energy because the mass defect of Y is greater than that of either W or X.
  2. It cannot be deduced whether the reaction releases energy because the properties of Z are not known.
  3. The reaction requires an input of energy because the mass defect of the products is greater than the sum of the mass defects of the reactants.
  4. Energy is released by the reaction because the binding energy of the products is greater than the sum of the binding energies of the reactants.
Show Answers Only

`D`

Show Worked Solution

Energy required to break W and  X  into their constituent nucleons:

 → 2.22 `+` 8.48 `=` 10.70 MeV

Energy released in the formation of the products is given by the sum of binding energies of  Y  and  Z:

 → 28.30 MeV ` +` binding energy of Z
   
As this is greater than the sum of binding energies of the reactants regardless of the binding energy of  Z , there is a net release of energy in the reaction.

`=>D`


♦ Mean mark 31%.

PHYSICS, M8 2020 HSC 11 MC

Consider the following nuclear reaction.

`\ _(3)^(6)text{Li} +\ _(0)^(1)text{n}rarr \ _(2)^(4)text{He} +\ _(1)^(3)text{H}`

The mass of the reactants is 7.023787704 `u` and the mass of the products is 7.018652532 `u`.

What type of reaction is this?

  1. A fusion reaction in which energy is released
  2. A fusion reaction that requires an input of energy
  3. A transmutation reaction in which energy is released
  4. A transmutation reaction that requires an input of energy
Show Answers Only

`C`

Show Worked Solution

The products contain multiple nuclei

→ The reaction is not a fusion reaction.

The mass of the products is less than the mass of the reactants

→ Energy is released.

`=>C`