When an electron jumps from a level `n = 4` to` n = 1`, the momentum of the recoiled hydrogen atom will be

Options

A. `6.5 xx 10^(-27) kg m s^(-1)`
B. `22.75 xx 10^(-19) kg m s^(-1)`
C. `13.6 xx 10^(-27) kg m s^(-1)`
D. zero

Explanation

To solve the problem of finding the momentum of the recoiled hydrogen atom when an electron jumps from the level \( n = 4 \) to \( n = 1 \), we will follow these steps:

### Step 1: Understand the process
When an electron transitions from a higher energy level to a lower energy level, it emits a photon. The hydrogen atom recoils due to the conservation of momentum.

### Step 2: Use the conservation of momentum
The momentum of the recoiled hydrogen atom will be equal in magnitude and opposite in direction to the momentum of the emitted photon. Therefore, we can express the momentum of the recoiled hydrogen atom as:
\[ p_{\text{H}} = -p_{\text{photon}} \]

### Step 3: Calculate the momentum of the emitted photon
The momentum of a photon can be calculated using the formula:
\[ p_{\text{photon}} = \frac{E}{c} \]
where \( E \) is the energy of the photon and \( c \) is the speed of light.

### Step 4: Calculate the energy of the photon
The energy of the emitted photon can be calculated using the Rydberg formula:
\[ E = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R_H \) is the Rydberg constant (\( R_H \approx 13.6 \, \text{eV} \)), \( n_f \) is the final energy level (1 in this case), and \( n_i \) is the initial energy level (4 in this case).

Substituting the values:
\[ E = 13.6 \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = 13.6 \left( 1 - \frac{1}{16} \right) = 13.6 \left( \frac{15}{16} \right) = 12.75 \, \text{eV} \]

### Step 5: Convert energy to joules
To convert the energy from electron volts to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = 12.75 \times 1.6 \times 10^{-19} \, \text{J} = 2.04 \times 10^{-18} \, \text{J} \]

### Step 6: Calculate the momentum of the photon
Now we can calculate the momentum of the photon:
\[ p_{\text{photon}} = \frac{E}{c} = \frac{2.04 \times 10^{-18}}{3 \times 10^8} \approx 6.8 \times 10^{-27} \, \text{kg m/s} \]

### Step 7: Find the momentum of the recoiled hydrogen atom
Since the momentum of the recoiled hydrogen atom is equal in magnitude to the momentum of the photon:
\[ p_{\text{H}} = 6.8 \times 10^{-27} \, \text{kg m/s} \]

### Final Answer
The momentum of the recoiled hydrogen atom is approximately:
\[ p_{\text{H}} \approx 6.8 \times 10^{-27} \, \text{kg m/s} \]