An electron in `H` atom makes a transition from `n = 3 to n = 1`. The recoil momentum of the `H` atom will be

Options

A. `6.45 xx10^(-27) Ns`
B. `6.8xx10^(-27)Ns`
C. `6.45xx10^(-24)Ns`
D. `6.8xx10^(-24)Ns`

Answer

To find the recoil momentum of the hydrogen atom when an electron transitions from \( n = 3 \) to \( n = 1 \), we can follow these steps:

### Step 1: Understand the Transition
When an electron transitions from a higher energy level to a lower energy level, it emits a photon. The momentum of the emitted photon will be equal and opposite to the recoil momentum of the hydrogen atom due to conservation of momentum.

### Step 2: Calculate the Wavelength of the Emitted Photon
The wavelength (\( \lambda \)) of the emitted photon can be calculated using the Rydberg formula for hydrogen:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( R \) is the Rydberg constant, approximately \( 1.1 \times 10^7 \, \text{m}^{-1} \)
- \( n_1 = 1 \) (final state)
- \( n_2 = 3 \) (initial state)

Substituting the values:

\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \left( \frac{8}{9} \right) \]

### Step 3: Substitute the Rydberg Constant
Now substituting \( R \):

\[ \frac{1}{\lambda} = 1.1 \times 10^7 \left( \frac{8}{9} \right) = \frac{8.8 \times 10^7}{9} \approx 9.78 \times 10^6 \, \text{m}^{-1} \]

### Step 4: Calculate the Wavelength
Now, take the reciprocal to find \( \lambda \):

\[ \lambda = \frac{1}{9.78 \times 10^6} \approx 1.02 \times 10^{-7} \, \text{m} \]

### Step 5: Calculate the Momentum of the Photon
The momentum (\( p \)) of the photon can be calculated using the formula:

\[ p = \frac{h}{\lambda} \]

Where \( h \) is Planck's constant, approximately \( 6.63 \times 10^{-34} \, \text{Js} \).

Substituting the values:

\[ p = \frac{6.63 \times 10^{-34}}{1.02 \times 10^{-7}} \approx 6.50 \times 10^{-27} \, \text{kg m/s} \]

### Step 6: Recoil Momentum of the Hydrogen Atom
By conservation of momentum, the recoil momentum of the hydrogen atom will be equal in magnitude to the momentum of the emitted photon:

\[ p_{\text{recoil}} = 6.50 \times 10^{-27} \, \text{kg m/s} \]

### Conclusion
Thus, the recoil momentum of the hydrogen atom when the electron transitions from \( n = 3 \) to \( n = 1 \) is approximately \( 6.50 \times 10^{-27} \, \text{kg m/s} \).

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