A nucleus of mass M emits an X-ray photon of frequency v. Energy lost by the nucleus is given as

Options

A. hv
B. (h^2v^2)/(2MC^2)`
C. `hv(1 - (hv)/(2MC^2))`
D. `hv(1 + (hv)/(2MC^2))`

Answer

To solve the problem of determining the energy lost by a nucleus that emits an X-ray photon of frequency \( \nu \), we can follow these steps:

### Step 1: Understand the Energy Loss
When a nucleus emits a photon, it loses energy. The total energy lost by the nucleus consists of two parts:
1. The kinetic energy of the nucleus after emission.
2. The energy of the emitted photon.

### Step 2: Write the Energy of the Photon
The energy \( E \) of a photon can be expressed using Planck's equation:
\[ E = h \nu \]
where \( h \) is Planck's constant and \( \nu \) is the frequency of the emitted photon.

### Step 3: Write the Kinetic Energy of the Nucleus
The kinetic energy \( K \) of the nucleus after emitting the photon can be expressed in terms of its momentum \( p \):
\[ K = \frac{p^2}{2M} \]
where \( M \) is the mass of the nucleus.

### Step 4: Relate Momentum to Frequency
From the theory of photons, we know that the momentum \( p \) of the photon can be related to its frequency:
\[ p = \frac{h \nu}{c} \]
where \( c \) is the speed of light.

### Step 5: Substitute Momentum into Kinetic Energy
Now, substituting the expression for momentum into the kinetic energy formula, we get:
\[ K = \frac{(h \nu / c)^2}{2M} = \frac{h^2 \nu^2}{2Mc^2} \]

### Step 6: Total Energy Lost by the Nucleus
The total energy lost by the nucleus when it emits the photon is the sum of the kinetic energy of the nucleus and the energy of the photon:
\[ \text{Energy lost} = K + E = \frac{h^2 \nu^2}{2Mc^2} + h \nu \]

### Step 7: Factor the Expression
We can factor \( h \nu \) out of the expression:
\[ \text{Energy lost} = h \nu \left(1 + \frac{h \nu}{2Mc^2}\right) \]

### Final Answer
Thus, the energy lost by the nucleus upon emitting the X-ray photon of frequency \( \nu \) is:
\[ \text{Energy lost} = h \nu \left(1 + \frac{h \nu}{2Mc^2}\right) \]