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 `10V`, the value of `n` is

Options

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

Explanation

To solve the problem, we need to find the value of \( n \) for the excited state from which the electron in the hydrogen atom jumps to the ground state. We will use the principles of energy conservation and the Bohr model of the hydrogen atom.

### Step-by-Step Solution:

1. **Identify the Energy of the Photon Emitted:**
The energy of the photon emitted when the electron jumps from an excited state \( n \) to the ground state (which is \( n=1 \)) can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the emitted photon.

2. **Relate Photon Energy to Work Function and Kinetic Energy:**
The energy of the emitted photon is also related to the work function \( \phi \) of the photo-sensitive material and the kinetic energy of the emitted photoelectrons. The stopping potential \( V_0 \) gives the kinetic energy of the photoelectrons as:
\[ KE = eV_0 \]
Therefore, we can write:
\[ E = \phi + KE \]
Substituting the values given in the problem:
\[ E = 2.75 \, \text{eV} + 10 \, \text{eV} = 12.75 \, \text{eV} \]

3. **Set Up the Equation Using Bohr's Model:**
According to the Bohr model, the energy levels of the hydrogen atom are given by:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
The energy difference when the electron transitions from state \( n \) to the ground state \( n=1 \) is:
\[ E = E_1 - E_n = -\frac{13.6}{1^2} - \left(-\frac{13.6}{n^2}\right) = 13.6 \left(1 - \frac{1}{n^2}\right) \]

4. **Equate the Two Expressions for Energy:**
Now we equate the two expressions for energy:
\[ 12.75 = 13.6 \left(1 - \frac{1}{n^2}\right) \]

5. **Solve for \( n^2 \):**
Rearranging the equation gives:
\[ 1 - \frac{1}{n^2} = \frac{12.75}{13.6} \]
Calculating the right-hand side:
\[ \frac{12.75}{13.6} \approx 0.9375 \]
Thus,
\[ 1 - \frac{1}{n^2} = 0.9375 \]
This simplifies to:
\[ \frac{1}{n^2} = 1 - 0.9375 = 0.0625 \]
Therefore,
\[ n^2 = \frac{1}{0.0625} = 16 \]
Taking the square root gives:
\[ n = 4 \]

6. **Conclusion:**
The electron jumps from the excited state \( n = 4 \) to the ground state \( n = 1 \).

### Final Answer:
The value of \( n \) is \( 4 \).

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