A stationary hydrogen atom emits photn corresponding to the first line of Lyman series. If R is the Rydberg constant and M is the mass of the atom, then the velocity acquired by the atom is

Options

A. `(3RH)/(4M)`
B. `(4M)/(3RH)`
C. `(RH)/(4M)`
D. `(4M)/(RH)`

Explanation

To solve the problem of finding the velocity acquired by a stationary hydrogen atom that emits a photon corresponding to the first line of the Lyman series, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Wavelength for the First Line of the Lyman Series**:
The Lyman series corresponds to transitions where the electron falls to the n=1 level. The first line of the Lyman series corresponds to the transition from n=2 to n=1.
The formula for the wavelength (λ) in the Lyman series is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the first line of the Lyman series, \( n_1 = 1 \) and \( n_2 = 2 \):
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Therefore, we have:
\[ \frac{1}{\lambda} = \frac{3R}{4} \]

2. **Relate Momentum and Wavelength**:
The momentum (P) of the emitted photon can be expressed in terms of its wavelength:
\[ P = \frac{h}{\lambda} \]
where \( h \) is Planck's constant.

3. **Substituting for Wavelength**:
From the previous step, we can substitute for \( \lambda \):
\[ P = \frac{h}{\lambda} = \frac{h}{\frac{4}{3R}} = \frac{3hR}{4} \]

4. **Conservation of Momentum**:
When the hydrogen atom emits a photon, it recoils due to conservation of momentum. The momentum of the atom (mass \( M \) and velocity \( v \)) must equal the momentum of the emitted photon:
\[ Mv = \frac{3hR}{4} \]

5. **Solving for Velocity**:
Rearranging the equation to solve for \( v \):
\[ v = \frac{3hR}{4M} \]

### Final Result:
The velocity acquired by the hydrogen atom after emitting the photon corresponding to the first line of the Lyman series is:
\[ v = \frac{3hR}{4M} \]