If the frequency of light in a photoelectric experiment is doubled the stopping potential will

Options

A. be doubled
B. be halved
C. become more than double
D. becomes less than double

Explanation

To solve the problem, we will use Einstein's photoelectric equation, which relates the stopping potential to the frequency of incident light and the work function of the material.

### Step-by-Step Solution:

1. **Understand the Photoelectric Equation**:
The photoelectric equation is given by:
\[ E \cdot V_0 = h \nu - \phi_0 \]
where:
- \(E\) is the charge of an electron,
- \(V_0\) is the stopping potential,
- \(h\) is Planck's constant,
- \(\nu\) is the frequency of the incident light,
- \(\phi_0\) is the work function of the material.

2. **Initial Condition**:
Let the initial frequency of light be \(\nu\). The stopping potential corresponding to this frequency is \(V_0\). Thus, we can write:
\[ E \cdot V_0 = h \nu - \phi_0 \quad \text{(Equation 1)} \]

3. **Doubling the Frequency**:
Now, if the frequency of light is doubled, we have:
\[ \nu' = 2\nu \]
The new stopping potential will be \(V_0'\). According to the photoelectric equation, we can write:
\[ E \cdot V_0' = h (2\nu) - \phi_0 \quad \text{(Equation 2)} \]

4. **Substituting and Rearranging**:
From Equation 2, we have:
\[ E \cdot V_0' = 2h \nu - \phi_0 \]
Now, we can express \(V_0'\) in terms of \(V_0\):
\[ E \cdot V_0' = 2(h \nu - \phi_0) + \phi_0 \]
We can substitute \(E \cdot V_0\) from Equation 1 into this equation:
\[ E \cdot V_0' = 2(E \cdot V_0) + \phi_0 \]

5. **Dividing the Equations**:
Now, we can divide Equation 2 by Equation 1:
\[ \frac{E \cdot V_0'}{E \cdot V_0} = \frac{2h \nu - \phi_0}{h \nu - \phi_0} \]
Simplifying this gives:
\[ \frac{V_0'}{V_0} = \frac{2h \nu - \phi_0}{h \nu - \phi_0} \]

6. **Analyzing the Result**:
We can simplify this further. The right-hand side can be expressed as:
\[ \frac{V_0'}{V_0} = 2 + \frac{\phi_0}{h \nu - \phi_0} \]
Since \(\phi_0\) is positive and \(h \nu > \phi_0\) (for photoelectric emission to occur), the term \(\frac{\phi_0}{h \nu - \phi_0}\) is also positive.

7. **Conclusion**:
Therefore, we conclude that:
\[ V_0' > 2V_0 \]
This means that the stopping potential when the frequency is doubled is greater than double the original stopping potential.

### Final Answer:
The stopping potential will be greater than double when the frequency of light is doubled.