Question

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 236. The total energy released in this process is

• A. \( 58E \)
• B. \( 59E \)
• C. \( \frac{58E}{59} \)
• D. \( \frac{59E}{58} \) \bigskip

Hint:

The energy released in the decay process depends on the relative masses involved and the type of radiation emitted.

Answer 1

The Correct Option is D

The energy released in the decay process involves the conversion of mass into energy according to Einstein's relation \( E = mc^2 \). Given the mass number 236 and the energy \( E \) of the alpha particle, the ratio can be calculated based on the energy equivalence and decay mechanism. 

 Step 1: The total energy released during the process of an $\alpha$-particle decay can be related to the ratio of mass numbers and energy based on decay theory. Using the formula derived from the decay laws, we calculate the total energy as: \[ E_{total} = \frac{59E}{58} \] Thus, the correct answer is \( \frac{59E}{58} \).