An electron in the hydrogen atom jumps from excited state `n` to the ground state. The wavelength so emitted illuminates a photo-sensitive material having work function `2.75eV`. If the stopping potential of the photoelectron is `10eV`, the value of `n` is

Options

A. `3`
B. `4`
C. `5`
D. `2`

Explanation

To solve the problem step by step, we will follow the outlined reasoning from the video transcript:

### Step 1: Understand the Energy Conservation
When an electron in a hydrogen atom transitions from an excited state (n) to the ground state (n=1), it emits energy in the form of a photon. The energy of the emitted photon must equal the work done to overcome the work function of the photo-sensitive material plus the kinetic energy of the emitted photoelectron.

### Step 2: Write the Energy Equation
The energy of the emitted photon can be expressed as:
\[ E = \text{Work Function} + \text{Kinetic Energy} \]
Where:
- Work Function \( \phi = 2.75 \, \text{eV} \)
- Stopping Potential \( V_s = 10 \, \text{eV} \)
- Kinetic Energy \( KE = eV_s = 10 \, \text{eV} \)

Thus, the total energy emitted is:
\[ E = 2.75 \, \text{eV} + 10 \, \text{eV} = 12.75 \, \text{eV} \]

### Step 3: Relate Emitted Energy to Photon Energy
The energy of the photon emitted during the transition can also be expressed using the formula:
\[ E = \frac{hc}{\lambda} \]
However, for hydrogen transitions, we can use the Rydberg formula:
\[ E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \( n_1 = 1 \) (ground state)
- \( n_2 = n \) (excited state)

### Step 4: Set Up the Equation
From the above, we can set up the equation:
\[ 12.75 \, \text{eV} = 13.6 \, \text{eV} \left( \frac{1}{1^2} - \frac{1}{n^2} \right) \]

### Step 5: Solve for \( n \)
Rearranging gives:
\[ 12.75 = 13.6 \left( 1 - \frac{1}{n^2} \right) \]
Dividing both sides by 13.6:
\[ \frac{12.75}{13.6} = 1 - \frac{1}{n^2} \]
Calculating the left side:
\[ 0.9375 = 1 - \frac{1}{n^2} \]
Rearranging gives:
\[ \frac{1}{n^2} = 1 - 0.9375 = 0.0625 \]
Taking the reciprocal:
\[ n^2 = \frac{1}{0.0625} = 16 \]
Thus:
\[ n = 4 \]

### Conclusion
The value of \( n \) is 4, meaning the electron was in the fourth excited state before transitioning to the ground state.