The electron in the hydrogen atom jumps form excited state (n=3) to its ground state (n=1) and the photons thus emitted irradiate a photosenstive 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. `5.1V`
B. `12.1V`
C. `17.2V`
D. `7V`

Explanation

To solve the problem, we need to calculate the stopping potential when an electron in a hydrogen atom transitions from the excited state (n=3) to the ground state (n=1). The energy of the emitted photon will be used to determine the stopping potential in relation to the work function of the photosensitive material.

### Step-by-Step Solution:

1. **Calculate the Energy of the Electron in the Excited State (n=3)**:
\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \text{ eV} \]

2. **Calculate the Energy of the Electron in the Ground State (n=1)**:
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV} \]

3. **Calculate the Energy of the Photon Emitted (ΔE)**:
The energy of the photon emitted when the electron transitions from n=3 to n=1 is given by:
\[ \Delta E = E_3 - E_1 = -1.51 \text{ eV} - (-13.6 \text{ eV}) = 12.1 \text{ eV} \]

4. **Determine the Maximum Kinetic Energy of the Emitted Photon**:
The maximum kinetic energy (K.E.) of the emitted photon can be expressed as:
\[ K.E. = \Delta E - \text{Work Function} \]
Given that the work function of the material is 5.1 eV, we can substitute:
\[ K.E. = 12.1 \text{ eV} - 5.1 \text{ eV} = 7.0 \text{ eV} \]

5. **Relate Kinetic Energy to Stopping Potential (V_s)**:
The stopping potential (V_s) is related to the maximum kinetic energy by the equation:
\[ K.E. = e \cdot V_s \]
Where \( e \) is the charge of the electron (1 eV corresponds to 1 volt). Therefore:
\[ V_s = K.E. = 7.0 \text{ volts} \]

### Final Answer:
The stopping potential is **7.0 volts**.

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