Question:
If \(E\) is the kinetic energy of the alpha particle emitted in the decay of a radioactive nucleus of mass number \(A\), then the disintegration energy released in the process is (Parent nucleus is at rest):
If \(E\) is the kinetic energy of the alpha particle emitted in the decay of a radioactive nucleus of mass number \(A\), then the disintegration energy released in the process is (Parent nucleus is at rest):
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In alpha decay, the daughter nucleus also gains kinetic energy. Therefore, the total disintegration energy is always greater than the alpha-particle kinetic energy.
Updated On: Oct 1, 2026
- \(\frac{(A+4)E}{A}\)
- \(\frac{AE}{A-4}\)
- \(\frac{(A-4)E}{A}\)
- \(\frac{4E}{A-4}\)
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The Correct Option is B
Solution and Explanation
Concept:
During alpha decay,
\[
Q=K_\alpha+K_D
\]
where \(K_\alpha\) and \(K_D\) are the kinetic energies of alpha particle and daughter nucleus respectively.
Using conservation of momentum,
\[
p_\alpha=p_D
\]
and
\[
\frac{K_\alpha}{K_D}
=
\frac{M_D}{M_\alpha}
\]
Step 1: Find kinetic energy of daughter nucleus. Mass of alpha particle \[ M_\alpha=4 \] Mass of daughter nucleus \[ M_D=A-4 \] Therefore, \[ \frac{E}{K_D} = \frac{A-4}{4} \] \[ K_D = \frac{4E}{A-4} \]
Step 2: Calculate disintegration energy. \[ Q = E+\frac{4E}{A-4} \] \[ Q = E\left(\frac{A-4+4}{A-4}\right) \] \[ Q = \frac{AE}{A-4} \] \[ \boxed{\frac{AE}{A-4}} \]
Step 1: Find kinetic energy of daughter nucleus. Mass of alpha particle \[ M_\alpha=4 \] Mass of daughter nucleus \[ M_D=A-4 \] Therefore, \[ \frac{E}{K_D} = \frac{A-4}{4} \] \[ K_D = \frac{4E}{A-4} \]
Step 2: Calculate disintegration energy. \[ Q = E+\frac{4E}{A-4} \] \[ Q = E\left(\frac{A-4+4}{A-4}\right) \] \[ Q = \frac{AE}{A-4} \] \[ \boxed{\frac{AE}{A-4}} \]
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