A nucleus of mass M emits an X-ray photon of frequency n. 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 energy lost by a nucleus when it emits an X-ray photon of frequency \( \nu \), we can follow these steps:

### Step 1: Understand the Energy Loss
When a nucleus emits an X-ray photon, it loses energy. This energy loss can be expressed as the sum of the energy carried away by the photon and the kinetic energy of the nucleus after the emission.

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

### Step 3: Consider the Kinetic Energy of the Nucleus
The kinetic energy \( KE \) of the nucleus after emitting the photon can be expressed as:
\[ KE = \frac{P^2}{2M} \]
where \( P \) is the momentum of the nucleus and \( M \) is the mass of the nucleus.

### Step 4: Relate Photon Momentum to Energy
The momentum \( P \) of the emitted photon can be expressed as:
\[ P = \frac{h}{\lambda} \]
Using the relationship between wavelength \( \lambda \) and frequency \( \nu \):
\[ \lambda = \frac{c}{\nu} \]
we can substitute to find:
\[ P = \frac{h \nu}{c} \]

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

### Step 6: Total Energy Loss
The total energy loss \( E_{loss} \) from the nucleus can be expressed as:
\[ E_{loss} = KE + E_{photon} = \frac{h^2 \nu^2}{2Mc^2} + h \nu \]

### Step 7: Factor Out Common Terms
We can factor out \( h \nu \):
\[ E_{loss} = h \nu \left(1 + \frac{h \nu}{2Mc^2}\right) \]

### Final Expression
Thus, the energy lost by the nucleus when it emits an X-ray photon is given by:
\[ E_{loss} = h \nu \left(1 + \frac{h \nu}{2Mc^2}\right) \]