Question

A nucleus of mass 218 amu free state decays to emit an $\alpha$ particle. Kinetic energy of the $\alpha$ particle emitted is 6.7MeV. The recoil energy (in MeV) of the daughter nucleus is:
A. 1
B. 0.5
C. 0.25
D. 0.125.

Answer 1

Use the law of conservation of momentum and find the relation between the momentums of the $\alpha$ particle and the daughter nucleus after the decay. Then use the relation between the kinetic energy and momentum of a particle.

Formula used:
${{P}_{i}}={{P}_{f}}$
$K=\dfrac{{{P}^{2}}}{2m}$

Complete step-by-step answer:
When the nucleus decays into its daughter nucleus and $\alpha$ particle, the net force on the system is zero. Therefore, the momentum of the system will be conserved. This means that the momentum of the system before the decay is equal to the momentum of the system of the decay.
Since before the decay the nucleus is at rest, the initial momentum of the system is zero.
i.e. ${{P}_{i}}=0$.
After the decay, the nucleus splits into a daughter nucleus and one $\alpha$ particle. The momentum of the $\alpha$ particle and the daughter nucleus are ${{P}_{1}}$ and ${{P}_{2}}$.
$\Rightarrow {{P}_{f}}={{P}_{1}}+{{P}_{2}}$
But, ${{P}_{i}}={{P}_{f}}$.
$\Rightarrow {{P}_{1}}=-{{P}_{2}}$.
If we consider only the magnitudes of the momentums then we get that ${{P}_{1}}={{P}_{2}}$.
The relation between the kinetic energy and momentum (magnitude) of a particle is given as $K=\dfrac{{{P}^{2}}}{2m}$, where m is the mass of the particle.
The kinetic energy and mass of $\alpha$ particle be ${{K}_{1}}$ and ${{m}_{1}}$.
The kinetic energy and mass of the daughter nucleus are ${{K}_{2}}$ and ${{m}_{2}}$.
Therefore,
${{K}_{1}}=\dfrac{P_{1}^{2}}{2{{m}_{1}}}$ ….. (i).
And
${{K}_{2}}=\dfrac{P_{2}^{2}}{2{{m}_{2}}}$ ….. (ii).
It is given that ${{K}_{1}}=6.7MeV$.
The mass of a $\alpha$ particle is 4amu.
Therefore, the mass of the daughter nucleus will be $218-4=216amu$.
$\Rightarrow {{m}_{1}}=4amu$
And
${{m}_{2}}=216amu$.
Substitute the respective values in (i) and (ii).
$\Rightarrow 6.7=\dfrac{P_{1}^{2}}{2(4)}$ ….. (iii).
And
${{K}_{2}}=\dfrac{P_{2}^{2}}{2(216)}$ …. (iv)
Now, divide (iv) by (iii).
$\Rightarrow \dfrac{{{K}_{2}}}{6.7}=\dfrac{\dfrac{P_{2}^{2}}{2(216)}}{\dfrac{P_{1}^{2}}{2(4)}}$
$\Rightarrow \dfrac{{{K}_{2}}}{6.7}=\dfrac{P_{2}^{2}}{P_{1}^{2}54}$
But ${{P}_{1}}={{P}_{2}}$.
$\Rightarrow \dfrac{{{K}_{2}}}{6.7}=\dfrac{1}{54}$
$\Rightarrow {{K}_{2}}=\dfrac{6.7}{54}\approx 0.125MeV$.

So, the correct answer is “Option D”.

Note: Note that when the net force on a system is zero, the net momentum of the system is constant. However, it is not true that the kinetic energy of the system will remain constant. In some cases, it may remain constant. Otherwise, it may convert into some other form of energy.