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 energy level \( n = 4 \) to \( n = 1 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Process**:
When an electron transitions from a higher energy level (n=4) to a lower energy level (n=1), it emits a photon. The momentum of the recoiled hydrogen atom will be equal in magnitude to the momentum of the emitted photon due to conservation of momentum.

2. **Photon Momentum**:
The momentum \( p \) of a photon can be expressed as:
\[ p = \frac{h}{\lambda} \]
where \( h \) is Planck's constant and \( \lambda \) is the wavelength of the emitted photon.

3. **Energy Levels of Hydrogen**:
The energy levels of the hydrogen atom are given by:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
The energy difference \( \Delta E \) between the two levels (n=4 and n=1) can be calculated as:
\[ \Delta E = E_1 - E_4 = -\frac{13.6}{1^2} - \left(-\frac{13.6}{4^2}\right) = -13.6 + \frac{13.6}{16} \]

4. **Calculating the Energy Difference**:
\[ \Delta E = -13.6 + 0.85 = -12.75 \, \text{eV} \]
Since we are interested in the magnitude of energy emitted:
\[ \Delta E = 12.75 \, \text{eV} \]

5. **Converting Energy to Joules**:
To convert the energy from eV to Joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \Delta E = 12.75 \times 1.6 \times 10^{-19} \, \text{J} = 2.04 \times 10^{-18} \, \text{J} \]

6. **Finding the Wavelength**:
The energy of a photon is also related to its wavelength by the equation:
\[ E = \frac{hc}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{hc}{E} \]
where \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \) and \( h = 6.63 \times 10^{-34} \, \text{Js} \).

7. **Calculating Wavelength**:
Substituting the values:
\[ \lambda = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{2.04 \times 10^{-18}} \approx 9.73 \times 10^{-8} \, \text{m} \]

8. **Finding Photon Momentum**:
Now we can find the momentum of the photon:
\[ p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{9.73 \times 10^{-8}} \approx 6.82 \times 10^{-27} \, \text{kg m/s} \]

9. **Momentum of Recoiled Hydrogen Atom**:
By conservation of momentum, the momentum of the recoiled hydrogen atom is equal to the momentum of the emitted photon:
\[ p_{\text{recoil}} = p_{\text{photon}} \approx 6.82 \times 10^{-27} \, \text{kg m/s} \]

### Final Answer:
The momentum of the recoiled hydrogen atom is approximately \( 6.82 \times 10^{-27} \, \text{kg m/s} \).