Question

An $\alpha$-particle is emitted from the decay of Americium (Am) at rest, i.e., $^{241}_{94}Am \Rightarrow ^{237}_{92}U + \alpha$. The rest masses of $^{241}_{94}Am$, $^{237}_{92}U$ and $\alpha$ are $224.544~\text{GeV}/c^{2}$, $220.811~\text{GeV}/c^{2}$ and $3.728~\text{GeV}/c^{2}$ respectively. What is the kinetic energy (in $\text{MeV}/c^{2}$, rounded off to two decimal places) of the $\alpha$-particle?

• A. 4.90
• B. 4.92
• C. 5.00
• D. 4.94

Hint:

In decay reactions, the kinetic energy of the emitted particle is the difference between the initial mass energy and the final mass energies of the products.

Answer 1

The Correct Option is A

- From the energy conservation law, the total energy before and after the decay must be equal. The kinetic energy of the $\alpha$-particle is given by: \[ E_{\alpha} = \left[ \text{Total energy of system} \right] - \left[ \text{Rest energy of }^{237}_{92}U \right] - \left[ \text{Rest energy of the $\alpha$-particle} \right]. \] - The total energy of the system is the rest mass energy of $^{241}_{94}Am$, which is 224.544 GeV/c\(^2\).
- The rest energy of $^{237}_{92}U$ is 220.811 GeV/c\(^2\) and of the $\alpha$-particle is 3.728 GeV/c\(^2\).
- Therefore, the kinetic energy of the $\alpha$-particle is: \[ E_{\alpha} = 224.544 - 220.811 - 3.728 = 4.90 \text{ MeV/c}^2. \] Thus, the correct answer is (A) 4.90.