A stationary nucleus of mass 24 amu emits a gamma photon. The energy of the emitted photon is 7 MeV. The recoil energy of the nucleus is

Options

A. (a) `2.2keV`
B. (b) `1.1 keV`
C. (c) `3.1 keV`
D. (d) `2.2 keV`

Explanation

To find the recoil energy of the nucleus after emitting a gamma photon, we can follow these steps:

### Step 1: Understand the conservation of momentum
When the nucleus emits a photon, both the photon and the nucleus must conserve momentum. Since the nucleus is initially stationary, the momentum of the emitted photon will equal the momentum of the recoiling nucleus.

### Step 2: Calculate the momentum of the emitted photon
The momentum \( P \) of a photon can be calculated using the formula:
\[ P = \frac{E}{c} \]
where \( E \) is the energy of the photon and \( c \) is the speed of light.

Given that the energy of the emitted photon \( E = 7 \, \text{MeV} \), we can convert this energy into joules for calculation purposes, but since we will be using it in the momentum formula, we can keep it in MeV.

### Step 3: Relate the momentum of the photon to the recoil momentum of the nucleus
Let \( m \) be the mass of the nucleus and \( v \) be the recoil velocity of the nucleus. The momentum of the nucleus after emission is given by:
\[ P_{\text{nucleus}} = mv \]
According to the conservation of momentum:
\[ \frac{E}{c} = mv \]

### Step 4: Solve for the recoil velocity \( v \)
From the momentum conservation equation, we can express the recoil velocity \( v \):
\[ v = \frac{E}{mc} \]

### Step 5: Calculate the recoil energy \( K \)
The kinetic energy (recoil energy) of the nucleus can be calculated using the formula:
\[ K = \frac{1}{2} mv^2 \]
Substituting \( v \) from the previous step:
\[ K = \frac{1}{2} m \left(\frac{E}{mc}\right)^2 \]
This simplifies to:
\[ K = \frac{E^2}{2mc^2} \]

### Step 6: Substitute the values
Now we need to substitute the values into the equation. We know:
- \( E = 7 \, \text{MeV} \)
- The mass of the nucleus \( m = 24 \, \text{amu} \)
- \( 1 \, \text{amu} = 931.5 \, \text{MeV}/c^2 \)

Thus, the mass in MeV/c² is:
\[ m = 24 \times 931.5 \, \text{MeV}/c^2 = 22356 \, \text{MeV}/c^2 \]

Now substituting these values into the kinetic energy equation:
\[ K = \frac{(7 \, \text{MeV})^2}{2 \times 22356 \, \text{MeV}/c^2} \]

### Step 7: Calculate the recoil energy
Calculating \( K \):
\[ K = \frac{49 \, \text{MeV}^2}{2 \times 22356 \, \text{MeV}/c^2} = \frac{49}{44712} \, \text{MeV} \approx 0.001095 \, \text{MeV} \approx 1.1 \, \text{keV} \]

### Final Answer
The recoil energy of the nucleus is approximately \( 1.1 \, \text{keV} \).

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