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PHYSICS, M8 EQ-Bank 7 MC

The following equation describes the natural decay process of uranium-238.

\({ }_{92}^{238} \text{U} \rightarrow{ }_{90}^{234}\text{Th} +{ }_2^4 \text{He}\)

Which row of the table describes the changes in total mass and total binding energy in the decay of uranium-238?

\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 \quad \textit{Total mass}\quad  \quad \rule[-1ex]{0pt}{0pt}& \textit{Total binding energy} \\
\hline
\rule{0pt}{2.5ex}\text{Decreases}\rule[-1ex]{0pt}{0pt}&\text{Increases}\\
\hline
\rule{0pt}{2.5ex}\text{Decreases}\rule[-1ex]{0pt}{0pt}& \text{Decreases}\\
\hline
\rule{0pt}{2.5ex}\text{Increases}\rule[-1ex]{0pt}{0pt}& \text{Increases} \\
\hline
\rule{0pt}{2.5ex}\text{Increases}\rule[-1ex]{0pt}{0pt}& \text{Decreases} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

→ Mass is converted into energy in this nuclear reaction, so total mass decreases.

→ When elements heavier than iron undergo nuclear fission, the daughter nuclei produced are more stable with a greater binding energy per nucleon. As there are the same number of nucleons before and after the decay, total binding energy increases.

\(\Rightarrow A\)

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 2020 HSC 16 MC

A model of the core of a nuclear fission reactor is shown.
 

When the reactor is operating normally, the moderator, control rods and coolant work in combination to maintain a controlled nuclear reaction in the fuel rods.

The moderator is a liquid which slows down neutrons to increase the rate of fission. The control rods absorb free neutrons. The coolant reduces the core temperature.

A fault causes some of the moderator to leak out of the core.

Which action would compensate for the effect of the loss of moderator?

  1. Withdraw the control rods from the core.
  2. Lower the control rods further into the core.
  3. Pump the coolant through the core at a faster rate.
  4. Reduce the temperature of the coolant before pumping it into the core.
Show Answers Only

`A`

Show Worked Solution

Loss of moderator leads to a reduction in the rate of fission. Withdrawing the control rods decreases the amount of neutrons absorbed, increasing the rate of fission.

`=>A`

♦ Mean mark 25%.