If the electron in hydrogen atom jumps from second Bohr orbit to ground state and difference between energies of the two states is radiated in the form of photons. If the work function of the material is `4.2 eV`, then stopping potential is
[Energy of electron in nth orbit `= - (13.6)/(n^(2)) eV`]

Options

A. 2 V
B. 4 V
C. 6 V
D. 8 V

Answer

To solve the problem, we need to calculate the stopping potential when an electron in a hydrogen atom transitions from the second Bohr orbit (n=2) to the ground state (n=1). We will also consider the work function of the material, which is given as 4.2 eV.

### Step-by-Step Solution:

1. **Calculate the energy of the electron in the second orbit (n=2)**:
The energy of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]
For n=2:
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV} \]

2. **Calculate the energy of the electron in the ground state (n=1)**:
For n=1:
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV} \]

3. **Calculate the energy difference (ΔE) between the two states**:
The energy difference when the electron jumps from n=2 to n=1 is:
\[ \Delta E = E_1 - E_2 = (-13.6) - (-3.4) = -13.6 + 3.4 = -10.2 \text{ eV} \]
The negative sign indicates that energy is released (radiated) as a photon.

4. **Determine the energy of the emitted photon**:
The energy of the emitted photon is:
\[ h\nu = 10.2 \text{ eV} \]

5. **Use the work function to find the stopping potential (V₀)**:
The stopping potential can be calculated using the equation:
\[ eV_0 = h\nu - \Phi \]
where \(\Phi\) is the work function (4.2 eV). Thus:
\[ V_0 = \frac{h\nu - \Phi}{e} \]
Substituting the values:
\[ V_0 = \frac{10.2 \text{ eV} - 4.2 \text{ eV}}{e} = \frac{6 \text{ eV}}{e} \]
Since we are looking for the stopping potential in volts, we can directly state:
\[ V_0 = 6 \text{ V} \]

### Final Answer:
The stopping potential is **6 V**.