A radioactive nucleus of mass `M` emits a photon of frequency `v` and the nucleus recoils. The recoil energy will be

Options

A. `h^(2)v^(2)//2 Mc^(2)`
B. zero
C. `hv`
D. `Mc^(2)-hv`

Answer

To find the recoil energy of a radioactive nucleus that emits a photon, we can follow these steps:

### Step 1: Understand the conservation of momentum
When the nucleus emits a photon, the momentum of the system must be conserved. The momentum of the emitted photon will be equal in magnitude and opposite in direction to the momentum gained by the recoiling nucleus.

### Step 2: Calculate the momentum of the 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. The energy of a photon is given by:
\[ E = h \nu \]
where \( h \) is Planck's constant and \( \nu \) is the frequency of the photon. Thus, the momentum of the photon becomes:
\[ P = \frac{h \nu}{c} \]

### Step 3: Relate the momentum to the recoil of the nucleus
Let \( M \) be the mass of the nucleus. The recoil momentum \( P_r \) of the nucleus will be equal to the momentum of the photon:
\[ P_r = P = \frac{h \nu}{c} \]

### Step 4: Calculate the recoil energy
The recoil energy \( E_r \) of the nucleus can be expressed in terms of its momentum:
\[ E_r = \frac{P_r^2}{2M} \]
Substituting the expression for \( P_r \):
\[ E_r = \frac{\left(\frac{h \nu}{c}\right)^2}{2M} \]

### Step 5: Simplify the expression
Now, simplify the expression for recoil energy:
\[ E_r = \frac{h^2 \nu^2}{2Mc^2} \]

### Final Answer
Thus, the recoil energy of the nucleus after emitting a photon of frequency \( \nu \) is given by:
\[ E_r = \frac{h^2 \nu^2}{2Mc^2} \]

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