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. become less than double

Answer

To solve the question regarding the effect of doubling the frequency of light on the stopping potential in a photoelectric experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**: The photoelectric effect is described by Einstein's photoelectric equation:
\[ h\nu = \phi_0 + K_{max} \]
where \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, \(\phi_0\) is the work function (threshold energy), and \(K_{max}\) is the maximum kinetic energy of the emitted electrons.

2. **Relate Kinetic Energy to Stopping Potential**: The maximum kinetic energy can also be expressed in terms of stopping potential \(V_0\):
\[ K_{max} = eV_0 \]
where \(e\) is the charge of the electron. Thus, we can rewrite the equation as:
\[ h\nu = \phi_0 + eV_0 \]

3. **Set Up the Equations for Two Cases**:
- **Case 1**: Let the initial frequency be \(\nu\) and the stopping potential be \(V_0\):
\[ eV_0 = h\nu - \phi_0 \quad \text{(Equation 1)} \]
- **Case 2**: When the frequency is doubled to \(2\nu\), let the new stopping potential be \(V_0'\):
\[ eV_0' = h(2\nu) - \phi_0 \quad \text{(Equation 2)} \]

4. **Express the New Stopping Potential**:
From Equation 2:
\[ eV_0' = 2h\nu - \phi_0 \]

5. **Divide the Two Equations**:
To find the relationship between \(V_0'\) and \(V_0\), divide Equation 2 by Equation 1:
\[ \frac{V_0'}{V_0} = \frac{2h\nu - \phi_0}{h\nu - \phi_0} \]

6. **Simplify the Expression**:
To simplify this, we can manipulate the numerator:
\[ \frac{V_0'}{V_0} = \frac{2h\nu - \phi_0}{h\nu - \phi_0} = \frac{2(h\nu - \phi_0) + \phi_0}{h\nu - \phi_0} \]
This can be rewritten as:
\[ \frac{V_0'}{V_0} = 2 + \frac{\phi_0}{h\nu - \phi_0} \]

7. **Analyze the Result**:
Since for photoelectric emission to occur, \(h\nu\) must be greater than \(\phi_0\) (i.e., \(h\nu > \phi_0\)), the term \(\frac{\phi_0}{h\nu - \phi_0}\) is positive. Therefore:
\[ \frac{V_0'}{V_0} > 2 \]

8. **Conclusion**:
This implies that the new stopping potential \(V_0'\) is greater than twice the original stopping potential \(V_0\). Thus, the correct answer is that the stopping potential will become more than double.

### Final Answer:
The stopping potential will become more than double when the frequency of light in a photoelectric experiment is doubled.

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