In a photoelectric effect experiment, stopping potential changes by 30 volt if we change frequency of the radiation. Then the magnitude of change in the frequency is : `(h=6xx10^(-34) J-s)`

Options

A. `4 xx 10^(-15) s^(-1)`
B. ` 8xx 10^(15) s^(-1)`
C. ` 10^(16) s^(-1)`
D. `18 xx10^(15) s^(-1)`

Answer

To solve the problem, we need to analyze the relationship between the stopping potential and the frequency of the incident radiation in the photoelectric effect.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The stopping potential \( V_0 \) in a photoelectric effect experiment is related to the frequency of the incident light by the equation:
\[ eV_0 = h\nu - \phi \]
where \( e \) is the charge of an electron, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident radiation, and \( \phi \) is the work function of the material.

2. **Change in Stopping Potential**:
According to the problem, the stopping potential changes by \( \Delta V_0 = 30 \) volts. This means:
\[ e \Delta V_0 = h \Delta \nu \]
where \( \Delta \nu \) is the change in frequency.

3. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ \Delta \nu = \frac{e \Delta V_0}{h} \]

4. **Substituting Known Values**:
We know:
- \( e = 1.6 \times 10^{-19} \) C (charge of an electron)
- \( h = 6 \times 10^{-34} \) J·s (Planck's constant)
- \( \Delta V_0 = 30 \) V

Substituting these values into the equation:
\[ \Delta \nu = \frac{(1.6 \times 10^{-19} \, \text{C})(30 \, \text{V})}{6 \times 10^{-34} \, \text{J·s}} \]

5. **Calculating the Change in Frequency**:
\[ \Delta \nu = \frac{4.8 \times 10^{-18}}{6 \times 10^{-34}} = 8 \times 10^{15} \, \text{Hz} \]

6. **Final Answer**:
The magnitude of the change in frequency is:
\[ \Delta \nu = 8 \times 10^{15} \, \text{Hz} \]

### Conclusion:
Thus, the answer to the question is \( 8 \times 10^{15} \, \text{Hz} \).

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