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PHYSICS, M8 2024 HSC 7 MC

A pure sample of polonium-210 undergoes alpha emission to produce the stable isotope lead-206.

The half-life of polonium-210 is 138 days.

At the end of 276 days, what is the ratio of polonium-210 atoms to lead-206 atoms in the sample?

  1. \(1 : 4\)
  2. \(1 : 3\)
  3. \(1 : 2\)
  4. \(1 : 1\)
Show Answers Only

\(B\)

Show Worked Solution
  • After two half-life’s of the polonium only one quarter of the sample of polonium-210 atoms will be left.
  • Therefore, three quarters of the polonium-210 atoms would have decayed into lead-206 atoms.
  • The ratio of polonium atoms to lead atoms is \(\dfrac{1}{4}:\dfrac{3}{4} = 1:3\)

\(\Rightarrow B\)

♦ Mean mark 45%.

PHYSICS, M8 2023 HSC 11 MC

The chart shows part of a nuclear decay series beginning with uranium.
 


 

Which option correctly identifies \(X\) and \(Y\) and the process by which each was produced?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\\
\textbf{}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\quad \quad \quad X\rule[-1ex]{0pt}{0pt}&\quad \quad \quad Y\\
\hline
\rule{0pt}{2.5ex}{ }_{\ \ 90}^{234}\text{Th}&{ }_{\ \ 91}^{234}\text{Pa}\\
\text{alpha decay} \quad \quad  \rule[-1ex]{0pt}{0pt}&\text{beta decay}\quad \quad \\
\hline
\rule{0pt}{2.5ex}{ }_{\ \ 90}^{234}\text{Th}& { }_{\ \ 91}^{234}\text{Pa}\\
\text{alpha decay}\quad \rule[-1ex]{0pt}{0pt}&\text{alpha decay}\quad \\
\hline
\rule{0pt}{2.5ex}{ }_{\ \ 91}^{234}\text{Pa}& { }_{\ \ 91}^{234}\text{Pa}\\
\text{beta decay}\quad \rule[-1ex]{0pt}{0pt}&\text{beta decay}\quad \\
\hline
\rule{0pt}{2.5ex}{ }_{\ \ 91}^{234}\text{Pa}& { }_{\ \ 90}^{234}\text{Th} \\
\text{beta decay}\quad \rule[-1ex]{0pt}{0pt}&\text{alpha decay}\quad \\
\hline
\end{array}
\end{align*} 

Show Answers Only

\(A\)

Show Worked Solution
  • During the decay process producing \(X\), the Atomic number decreases by 2 to 90 and the Mass number decreases by 4 to 234
  • Alpha decay producing  \({ }_{\ \ 90}^{234}\text{Th}\)
  • During the decay process producing \(Y\), the Atomic number increases by 1 to 91 and the Mass number does not change. 
  • Beta decay producing \({ }_{\ \ 91}^{234}\text{Pa}\)

\(\Rightarrow A\)

PHYSICS, M8 2023 HSC 4 MC

Caesium-137 has a half-life of 30 years.

What mass of caesium-137 will remain after 90 years, if the initial mass was 120 g?

  1. 4 g
  2. 15 g
  3. 40 g
  4. 60 g
Show Answers Only

\(B\)

Show Worked Solution

\(t_\frac{1}{2} = \text{30 years,}\ \ \lambda =\dfrac{\ln2}{30}\)

\(N=N_{0}e^{-\lambda t} =120e^{-\dfrac{\ln2}{30} \times 90} =15\ \text{g}\)

 \(\Rightarrow B\)

PHYSICS, M8 EQ-Bank 23

A patient is given an injection containing `6.0 × 10^(-18)` kg of radioactive technetium-99m which has a half-life of 6 hours.

How much remains undecayed when a scan is taken 3 hours later?   (3 marks)

Show Answers Only

`4.2 xx10^(-18)\ text{kg}`

Show Worked Solution

Calculating the decay constant:

   `lambda=(ln2)/(t_((1)/(2)))=(ln2)/(6)=0.1155\ text{hour}^(-1)`

Amount remaining undecayed:

`N` `=N_(0)e^(-lambdat)`  
  `=6.0 xx10^(-18) xxe^(-0.1155 xx3)`  
  `=4.2429… xx10^(-18)`  
  `=4.2 xx10^(-18)\ text{kg}`  

PHYSICS, M8 EQ-Bank 3 MC

A 5-gram sample of radioactive strontium-90 decayed over time. The graph shows the mass of strontium-90 remaining from the initial sample as a function of time.
 

What is the approximate value of the decay constant, in `\text{year}^(-1)`, for strontium-90?

  1. 0.006
  2. 0.011
  3. 0.014
  4. 0.025
Show Answers Only

`D`

Show Worked Solution
  • From the graph, the half life (mass = 2.5 g) is approximately 29 years.
  •    `lambda=(ln2)/(t_((1)/(2)))=(ln2)/(29)=0.0239\ text{year}^(-1)`

 `=>D`

PHYSICS M8 2022 HSC 24

The radioactive decay curve for americium-242 is shown.
 


 

  1. Use the graph to find the half-life of Am-242 and hence show that the decay constant, `\lambda`, is `0.043` `text{h}^(-1).`   (2 marks)
  1. Calculate how long it takes until the mass of Am-242 is 8 micrograms.   (2 marks)
Show Answers Only

a.   `0.043  text{h}^(-1)`

b.  `53.55\ text{h}`

Show Worked Solution

a.   The half life of Am-242 is 16 hours.

`lambda=(ln 2)/(t_(1//2))=(ln 2)/(16)=0.043  text{h}^(-1)`
 

b.    `N` `=N_(0)e^(-lambda t)`
     `8` `=80e^(-0.043 t)`
  `e^(-0.043 t)` `=8/80`
  `-0.043t` `=ln(1/10)`
     `t` `=((-1)/(0.043))ln ((1)/(10))`
    `=53.55\ text{h}`

PHYSICS, M8 2021 HSC 35

A spacecraft is powered by a radioisotope generator. Pu-238 in the generator undergoes alpha decay, releasing energy. The decay is shown with the mass of each species in atomic mass units, `u`

\begin{array} {ccccc}
\ce{^{238}Pu} & \rightarrow & \ce{^{234}U} & + & \alpha \\
238.0495\ u &  & 234.0409\ u &  & 4.0026\ u \end{array}

  1. Show that the energy released by one decay is  `9.0 × 10^(-13)` J.   (3 marks)
  1. At launch, the generator contains  `9.0 × 10^24`  atoms of Pu-238. The half-life of Pu-238 is 87.7 years.
  2. Calculate the total energy produced by the generator during the first ten years after launch.  (3 marks)
Show Answers Only
  1. `9.0xx10^(-13)\ \text{J}`
  2. `6.2xx10^(11)\ \text{J}`
Show Worked Solution

a.   `text{Find Energy released:}`

`Delta m` `=(234.0409+4.0026)-238.0495=-0.006\ u`  
`Delta m` `=0.006 xx1.661 xx10^(-27)=9.966 xx10^(-30)\ \text{kg}`  

  
`\text{Using}\ \ E=mc^2:`

`E_text{released}=9.966 xx10^(-30)xx(3xx10^(8))^(2)=9.0 xx10^(-13)\ \text{J}`

   

b.   `\text{Find total energy:}`

`lambda=(ln 2)/(t_((1)/(2)))=(ln 2)/(87.7)=0.0079\ text{year}^(-1)`

`N=N_(0)e^(-lambda t)=9xx10^(24)xxe^(-0.0079 xx10)=8.316 xx10^(24)`

`Delta N=9xx10^(24)-8.316 xx10^(24)=6.84 xx10^(23)`

`E=6.84 xx10^(23)xx9.0 xx10^(-13)=6.2 xx10^(11)\ \text{J}`


♦ Mean mark part (b) 51%.

PHYSICS, M8 2020 HSC 4 MC

The graph shows the mass of a radioactive isotope as a function of time.
 

What is the decay constant, in `text{years}^(-1)`, for this isotope?

  1. 0.0030
  2. 0.0069
  3. 2.0
  4. 100
Show Answers Only

`B`

Show Worked Solution
`t_((1)/(2))` `~~100` years  
`lambda` `=(ln 2)/(t_((1)/(2)))=(ln 2)/(100)=0.0069  text{years}^(-1)`  

 
`=>B`

PHYSICS, M8 2021 HSC 19 MC

Rh-106 is a metallic, beta-emitting radioisotope with a half-life of 30 seconds.

A sample of Rh-106 and an electrode are placed inside an evacuated chamber. They are connected to a galvanometer and a variable DC power supply.
 

A student measures the current, `I`, when the power supply is set to zero. They then measure the stopping voltage, `V_s`. The stopping voltage is the minimum voltage needed to prevent current flowing.

A few minutes later, these measurements are repeated.

How do the TWO sets of measurements compare?

  1. Only `I` changes.
  2. Only `V_s` changes.
  3. Both `I` and `V_s` change.
  4. Neither `I` nor `V_s` changes.
Show Answers Only

`A`

Show Worked Solution
  • Due to the short half life of the isotope, there will be significantly less present when the second reading is made.
  • Lower rate of emission of electrons → lower recorded current.
  • The energy of emitted electrons will remain the same → voltage will not change.

`=>A`


♦ Mean mark 29%.