A stationary hydrogen atom in the first excited state emits a photon. If the mass of the hydrogen atom is m and its ionization energy is E, then the recoil velocity acquired by the atom is [speed of light = c]

Options

A. `[sqrt((3E)/(2m)+c^(2))]-c`
B. `[sqrt((3E)/(4m)+c^(2))]-c`
C. `(3E)/(4mc)`
D. `(E )/(mc)`

Answer

To find the recoil velocity acquired by a stationary hydrogen atom in the first excited state after emitting a photon, we can follow these steps:

### Step 1: Determine the energy of the emitted photon
The energy of the photon emitted when the hydrogen atom transitions from the first excited state (n=2) to the ground state (n=1) can be calculated using the formula for the energy levels of hydrogen:

\[ E = -\frac{13.6 \, \text{eV}}{n^2} \]

The energy difference between the two states is given by:

\[ E_{photon} = E_1 - E_2 = \left(-\frac{13.6 \, \text{eV}}{1^2}\right) - \left(-\frac{13.6 \, \text{eV}}{2^2}\right) = 13.6 \, \text{eV} \left(1 - \frac{1}{4}\right) = 13.6 \, \text{eV} \cdot \frac{3}{4} = 10.2 \, \text{eV} \]

### Step 2: Relate the energy of the photon to its momentum
The momentum \( p \) of the photon can be expressed in terms of its energy \( E \) using the relation:

\[ p = \frac{E}{c} \]

Substituting the energy of the photon:

\[ p = \frac{10.2 \, \text{eV}}{c} \]

### Step 3: Apply conservation of momentum
Before the emission, the total momentum of the system (hydrogen atom) is zero since it is stationary. After the emission of the photon, the momentum of the hydrogen atom \( m v_r \) (where \( v_r \) is the recoil velocity) must equal the momentum of the emitted photon:

\[ m v_r = \frac{E_{photon}}{c} \]

### Step 4: Solve for the recoil velocity
Rearranging the equation gives:

\[ v_r = \frac{E_{photon}}{m c} \]

Substituting the value of \( E_{photon} \):

\[ v_r = \frac{10.2 \, \text{eV}}{m c} \]

### Step 5: Express in terms of ionization energy
Given that the ionization energy \( E \) of hydrogen is 13.6 eV, we can express \( v_r \) in terms of \( E \):

\[ v_r = \frac{10.2 \, \text{eV}}{m c} = \frac{10.2}{13.6} \cdot \frac{E}{m c} = \frac{3}{4} \cdot \frac{E}{m c} \]

### Final Result
Thus, the recoil velocity \( v_r \) acquired by the hydrogen atom is:

\[ v_r = \frac{3E}{4mc} \]