Q157 / 160 · TS EAMCET 2024 (11 May Shift 1) · Physics
An $\alpha$-particle of energy ' $E$ ' is liberated during the decay of a nucleus of mass number 236. The total energy released in this process is
- A$58 E$
- B$59 E$
- C$\frac{58 E}{59}$
- Correct answerD$\frac{59 E}{58}$
Solution
In decay of nucleus, $(\mathrm{KE})_{\alpha}=\mathrm{E}$
As, $\mathrm{KE}=\frac{\mathrm{P}^2}{2 \mathrm{~m}} \Rightarrow \mathrm{KE} \propto \frac{1}{\mathrm{~m}}$ $\therefore \frac{(\mathrm{KE})_{232}}{(\mathrm{KE})_{\alpha}}=\frac{4}{232}=\frac{1}{58} \cdot \therefore(\mathrm{KE})_{232}=\frac{\mathrm{E}}{58}$ $\therefore$ Total energy released in this process, $\mathrm{Q}=(\mathrm{KE})_{\propto}+(\mathrm{KE})_{232}=\mathrm{E}+\frac{\mathrm{E}}{58}=\frac{59 \mathrm{E}}{58}$
As, $\mathrm{KE}=\frac{\mathrm{P}^2}{2 \mathrm{~m}} \Rightarrow \mathrm{KE} \propto \frac{1}{\mathrm{~m}}$ $\therefore \frac{(\mathrm{KE})_{232}}{(\mathrm{KE})_{\alpha}}=\frac{4}{232}=\frac{1}{58} \cdot \therefore(\mathrm{KE})_{232}=\frac{\mathrm{E}}{58}$ $\therefore$ Total energy released in this process, $\mathrm{Q}=(\mathrm{KE})_{\propto}+(\mathrm{KE})_{232}=\mathrm{E}+\frac{\mathrm{E}}{58}=\frac{59 \mathrm{E}}{58}$
Hence, the correct answer is option: 'D'.