Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be `3.57V`. The threshold frequency of the material is

Options

A. `1.6xx10^(15) Hz`
B. `2.5xx10^(15) Hz`
C. `4xx10^(15) Hz`
D. `5xx10^(15)Hz`

Explanation

To solve the problem step by step, we will follow the concepts of energy transitions in hydrogen atoms and the photoelectric effect.

### Step-by-Step Solution:

1. **Identify the Energy Transition**:
The electron in a hydrogen atom jumps from the first excited state (n=2) to the ground state (n=1). The energy emitted during this transition can be calculated using the formula:
\[ E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Here, \( n_1 = 1 \) and \( n_2 = 2 \).

2. **Calculate the Energy Emitted**:
Substituting the values into the formula:
\[ E = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \left( \frac{3}{4} \right) = 10.2 \text{ eV} \]

3. **Use the Stopping Potential**:
The stopping potential \( V_s \) is given as \( 3.57 \text{ V} \). The kinetic energy (KE) of the emitted electrons can be expressed as:
\[ KE = e \cdot V_s = 3.57 \text{ eV} \]

4. **Relate the Total Energy to Work Function and Kinetic Energy**:
The total energy emitted (E) is equal to the work function (\( \phi \)) plus the kinetic energy (KE):
\[ E = \phi + KE \]
Substituting the known values:
\[ 10.2 \text{ eV} = \phi + 3.57 \text{ eV} \]

5. **Calculate the Work Function**:
Rearranging the equation to find the work function:
\[ \phi = 10.2 \text{ eV} - 3.57 \text{ eV} = 6.63 \text{ eV} \]

6. **Calculate the Threshold Frequency**:
The work function is related to the threshold frequency (\( \nu_0 \)) by the equation:
\[ \phi = h \nu_0 \]
Where \( h \) (Planck's constant) is approximately \( 4.14 \times 10^{-15} \text{ eV s} \). Rearranging gives:
\[ \nu_0 = \frac{\phi}{h} \]
Substituting the values:
\[ \nu_0 = \frac{6.63 \text{ eV}}{4.14 \times 10^{-15} \text{ eV s}} \approx 1.60 \times 10^{15} \text{ Hz} \]

### Final Answer:
The threshold frequency of the material is approximately \( 1.60 \times 10^{15} \text{ Hz} \).

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