A hydrogen atom emits a photon corresponding to an electron transition from `n = 5` to `n = 1`. The recoil speed of hydrogen atom is almost (mass of proton `~~1.6 xx 10^(-27) kg)`.

Options

A. `10 ms^(-1)`
B. `2 xx 10^(-2) ms^(-1)`
C. `4 ms^(-1)`
D. `8 xx 10^(2) ms^(-1)`

Explanation

To find the recoil speed of a hydrogen atom when it emits a photon corresponding to an electron transition from \( n = 5 \) to \( n = 1 \), we will follow these steps:

### Step 1: Calculate the energy of the emitted photon
The energy of the photon emitted during the transition can be calculated using the formula:

\[ E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( n_1 = 1 \)
- \( n_2 = 5 \)

Substituting the values:

\[ E = 13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{5^2} \right) = 13.6 \, \text{eV} \left( 1 - \frac{1}{25} \right) = 13.6 \, \text{eV} \left( \frac{24}{25} \right) \]

Calculating this gives:

\[ E = 13.6 \times 0.96 = 13.056 \, \text{eV} \]

### Step 2: Convert energy from eV to Joules
To convert the energy from electron volts to Joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

\[ E = 13.056 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.089 \times 10^{-18} \, \text{J} \]

### Step 3: Calculate the momentum of the emitted photon
The momentum \( p \) of a photon can be calculated using the relationship:

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

Where \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \):

\[ p = \frac{2.089 \times 10^{-18} \, \text{J}}{3 \times 10^8 \, \text{m/s}} = 6.963 \times 10^{-27} \, \text{kg m/s} \]

### Step 4: Apply conservation of momentum
Since the initial momentum of the hydrogen atom is zero, the momentum of the emitted photon will equal the momentum of the recoiling hydrogen atom:

\[ p_{\text{photon}} = p_{\text{atom}} \]

Let \( m \) be the mass of the hydrogen atom (mass of proton \( \approx 1.6 \times 10^{-27} \, \text{kg} \)) and \( v \) be the recoil speed of the hydrogen atom:

\[ p_{\text{atom}} = m \cdot v \]

Thus, we have:

\[ 6.963 \times 10^{-27} \, \text{kg m/s} = 1.6 \times 10^{-27} \, \text{kg} \cdot v \]

### Step 5: Solve for the recoil speed \( v \)
Rearranging the equation gives:

\[ v = \frac{6.963 \times 10^{-27} \, \text{kg m/s}}{1.6 \times 10^{-27} \, \text{kg}} = 4.352 \, \text{m/s} \]

### Final Answer:
The recoil speed of the hydrogen atom is approximately \( 4.352 \, \text{m/s} \).

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