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 solve the problem of determining 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: Understand the Energy Transition
The hydrogen atom transitions from the first excited state (n=2) to the ground state (n=1). The energy difference between these states corresponds to the energy of the emitted photon.

### Step 2: Calculate the Energy of the Photon
The energy of the emitted photon can be calculated using the formula for the energy levels of the hydrogen atom:
\[ E_n = -\frac{E}{n^2} \]
where \(E\) is the ionization energy of the hydrogen atom.

The energy of the photon emitted during the transition from n=2 to n=1 is:
\[ E_{\text{photon}} = E_2 - E_1 = \left(-\frac{E}{2^2}\right) - \left(-\frac{E}{1^2}\right) = -\frac{E}{4} + E = \frac{3E}{4} \]

### Step 3: Apply Conservation of Momentum
Since the hydrogen atom is initially at rest, the momentum before the emission of the photon is zero. After the emission, the momentum of the photon must equal the momentum of the recoiling hydrogen atom:
\[ p_{\text{photon}} = p_{\text{atom}} \]
The momentum of the photon is given by:
\[ p_{\text{photon}} = \frac{E_{\text{photon}}}{c} = \frac{3E/4}{c} \]
The momentum of the recoiling hydrogen atom is:
\[ p_{\text{atom}} = m v_r \]
where \(v_r\) is the recoil velocity of the hydrogen atom.

Setting these equal gives:
\[ \frac{3E}{4c} = mv_r \]

### Step 4: Solve for Recoil Velocity
Rearranging the equation for \(v_r\):
\[ v_r = \frac{3E}{4mc} \]

### Step 5: Final Expression
Thus, the recoil velocity acquired by the hydrogen atom after emitting the photon is:
\[ v_r = \frac{3E}{4mc} \]