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PHYSICS, M8 2025 HSC 16 MC

A neutron is absorbed by a nucleus, \(X\).

The resulting nucleus undergoes alpha decay, producing lithium-7.

What is nucleus \(X\) ?

  1. Boron-10
  2. Boron-11
  3. Lithium-6
  4. Lithium-10
Show Answers Only

\(A\)

Show Worked Solution
  • The alpha decay described can be expresses as:
  •      \(\ce{^1_0n + ^{10}_5Bu \rightarrow ^7_3 Li + ^4_2 He}\)

\(\Rightarrow A\)

PHYSICS, M8 2025 HSC 10 MC

The diagram shows four lines, \(W, X, Y\) and \(Z\), depicting radioactivity varying with time.
 

Which of the four lines is consistent with a decay graph with the smallest decay constant \((\lambda)\) ?

  1. \(W\)
  2. \(X\)
  3. \(Y\)
  4. \(Z\)
Show Answers Only

\(C\)

Show Worked Solution
  • A radioactive decay graph is shaped exponentially and is not linear (eliminate \(W\) and \(Z\).
  • Since the decay constant \(=\dfrac{\lambda}{t_{\frac{1}{2}}}\), and \(Y\) has a higher half-life than \(X\) (its activity graaph is higher), \(Y\) must have the smaller decay constant.

\(\Rightarrow C\)

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%.