A `gamma` photon of momentum P is emitted by a radioactive nucleus of mass M . The total energy released in this process will be (Assume nuclei is free to recoil) [c is the speed of light ]

Options

A. Pc
B. `(P^2)/(2M)`
C. `P(c+P/(2M))`
D. `P(c+P/(4M))`

Answer

To solve the problem of finding the total energy released when a gamma photon of momentum \( P \) is emitted by a radioactive nucleus of mass \( M \), we can follow these steps:

### Step 1: Understand the Energy of the Gamma Photon
The energy \( E \) of a photon can be expressed in terms of its momentum \( P \) and the speed of light \( c \):
\[ E = Pc \]
This equation shows that the energy of the photon is directly proportional to its momentum.

**Hint:** Remember that the energy of a photon is related to its momentum and the speed of light.

### Step 2: Calculate the Kinetic Energy of the Recoiling Nucleus
When the gamma photon is emitted, the nucleus will recoil due to conservation of momentum. The kinetic energy \( KE \) of the recoiling nucleus can be expressed as:
\[ KE = \frac{P^2}{2M} \]
where \( M \) is the mass of the nucleus and \( P \) is the momentum of the emitted photon.

**Hint:** Use the formula for kinetic energy, which involves the mass and the square of the velocity.

### Step 3: Total Energy Released
The total energy released in the process is the sum of the energy of the emitted photon and the kinetic energy of the recoiling nucleus:
\[ \text{Total Energy} = E + KE = Pc + \frac{P^2}{2M} \]

**Hint:** The total energy is the sum of the energy of the photon and the kinetic energy of the recoiling nucleus.

### Step 4: Factor Out \( P \)
To simplify the expression for total energy, we can factor out \( P \):
\[ \text{Total Energy} = P \left( c + \frac{P}{2M} \right) \]

**Hint:** Factoring can help simplify complex expressions and make them easier to analyze.

### Conclusion
Thus, the total energy released in this process when a gamma photon of momentum \( P \) is emitted by a radioactive nucleus of mass \( M \) is:
\[ \text{Total Energy} = Pc + \frac{P^2}{2M} \]

### Final Answer
The total energy released in this process will be \( Pc + \frac{P^2}{2M} \).