The electron in the hydrogen atom jumps from excited state (n=3) to its ground state (n=1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1eV, the stopping potential is estimated to be: (The energy of the electron in nth state is `E_(n)=-13.6//n^(2)eV`)

Options

A. `5.1 V`
B. `12.1 V`
C. `17.2 V`
D. `7 V`

Answer

To solve the problem step by step, we will first calculate the energy of the photon emitted when the electron in the hydrogen atom transitions from the excited state (n=3) to the ground state (n=1). Then, we will find the stopping potential based on the work function of the photosensitive material.

### Step 1: Calculate the energy of the photon emitted

The energy of an electron in the nth state of a hydrogen atom is given by the formula:

\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]

1. **Calculate \(E_3\)** (energy at n=3):
\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \text{ eV} \]

2. **Calculate \(E_1\)** (energy at n=1):
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV} \]

3. **Calculate the energy of the photon emitted (\(E_{photon}\))**:
\[ E_{photon} = E_3 - E_1 = \left(-\frac{13.6}{9}\right) - (-13.6) = -1.51 + 13.6 = 12.1 \text{ eV} \]

### Step 2: Calculate the stopping potential

The stopping potential (\(V_0\)) can be calculated using the formula:

\[ V_0 = \frac{E_{photon} - \text{Work Function}}{e} \]

Where:
- \(E_{photon} = 12.1 \text{ eV}\)
- Work Function = \(5.1 \text{ eV}\)

1. **Substituting the values**:
\[ V_0 = E_{photon} - \text{Work Function} = 12.1 \text{ eV} - 5.1 \text{ eV} = 7.0 \text{ eV} \]

### Final Answer

The stopping potential is:
\[ V_0 = 7 \text{ V} \]