The recoil speed of a hydrogen atom after it emits a photon in going from `n=5`, state to `n=1` state is (in `ms^(-1)`)

Options

A. 4.718
B. 7.418
C. 4.178
D. 7.148

Explanation

To find the recoil speed of a hydrogen atom after it emits a photon while transitioning from the n=5 state to the n=1 state, we can follow these steps:

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

\[ E = R_H \cdot h \cdot c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( R_H \) is the Rydberg constant (\( 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \))
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \))
- \( n_1 = 1 \) (final state)
- \( n_2 = 5 \) (initial state)

Substituting the values:

\[ E = R_H \cdot h \cdot c \left( \frac{1}{1^2} - \frac{1}{5^2} \right) = R_H \cdot h \cdot c \left( 1 - \frac{1}{25} \right) = R_H \cdot h \cdot c \left( \frac{24}{25} \right) \]

### Step 2: Calculate the momentum of the emitted photon
The momentum \( p \) of a photon can be calculated using the relation:

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

Substituting the expression for energy from Step 1:

\[ p = \frac{R_H \cdot h \cdot c \cdot \left( \frac{24}{25} \right)}{c} = R_H \cdot h \cdot \left( \frac{24}{25} \right) \]

### Step 3: Relate photon momentum to recoil momentum of the hydrogen atom
According to the law of conservation of momentum, the momentum of the emitted photon is equal to the momentum of the recoiling hydrogen atom:

\[ p = m \cdot v \]

Where:
- \( m \) is the mass of the hydrogen atom (approximately \( 1.67 \times 10^{-27} \, \text{kg} \))
- \( v \) is the recoil speed we want to find.

Thus, we can write:

\[ R_H \cdot h \cdot \left( \frac{24}{25} \right) = m \cdot v \]

### Step 4: Solve for the recoil speed \( v \)
Rearranging the equation gives:

\[ v = \frac{R_H \cdot h \cdot \left( \frac{24}{25} \right)}{m} \]

Substituting the known values:

\[ v = \frac{(1.097 \times 10^7) \cdot (6.63 \times 10^{-34}) \cdot \left( \frac{24}{25} \right)}{1.67 \times 10^{-27}} \]

### Step 5: Calculate the recoil speed
Calculating the above expression:

1. Calculate \( R_H \cdot h \):
\[ R_H \cdot h = 1.097 \times 10^7 \times 6.63 \times 10^{-34} \approx 7.29 \times 10^{-27} \]

2. Now multiply by \( \frac{24}{25} \):
\[ 7.29 \times 10^{-27} \cdot \frac{24}{25} \approx 7.01 \times 10^{-27} \]

3. Finally, divide by the mass of the hydrogen atom:
\[ v \approx \frac{7.01 \times 10^{-27}}{1.67 \times 10^{-27}} \approx 4.19 \, \text{m/s} \]

Thus, the recoil speed of the hydrogen atom after emitting the photon is approximately \( 4.19 \, \text{m/s} \).

### Final Answer:
The recoil speed of the hydrogen atom is approximately \( 4.19 \, \text{m/s} \).

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