Question

A radioactive nucleus of mass $M$ emits a photon of frequency $\upsilon$ and the nucleus recoils. The recoil energy will be

• A. $Mc^2-h\upsilon$
• B. $h^2\upsilon^2 /2 Mc^2$
• C. zero
• D. h$\upsilon$

Answer 1

The Correct Option is B

The problem involves the physics of a radioactive nucleus emitting a photon and experiencing recoil, a situation that can be analyzed using the principles of conservation of momentum and energy.

Concept: When a nucleus emits a photon, it undergoes a recoil to conserve momentum. The photon carries momentum away, and by Newton's Third Law, the nucleus moves in the opposite direction.

Let’s break down the solution:

1. Conservation of Momentum: When the nucleus emits a photon of frequency $\upsilon$, it carries momentum $p = \frac{h\upsilon}{c}$, where $h$ is Planck’s constant and $c$ is the speed of light.

2. The nucleus must recoil with momentum $p_{\text{recoil}} = \frac{h\upsilon}{c}$ to conserve momentum.

3. Recoil Energy: The kinetic energy (recoil energy) of the nucleus can be expressed using the formula for kinetic energy:

   $\text{KE} = \frac{p^2}{2M}$

   Substitute $p = \frac{h\upsilon}{c}$:

   $E_{\text{recoil}} = \frac{(\frac{h\upsilon}{c})^2}{2M} = \frac{h^2\upsilon^2}{2Mc^2}$

4. Thus, the recoil energy of the nucleus is $E_{\text{recoil}} = \frac{h^2\upsilon^2}{2Mc^2}$, which corresponds to the correct answer.

Conclusion: Among the given options, the correct answer is $h^2\upsilon^2 /2 Mc^2$. This option describes the energy associated with the recoil of the nucleus, aligning with the conservation laws and kinematic principles applicable to the emission of a photon by a nucleus.