In a photoelectric effect measurement, the stopping potential for a given metal is found to be `V_(0)` volt, when radiation of wavelength `lamda_(0)` is used. If radiation of wavelength `2lamda_(0)` is used with the same metal, then the stopping potential (in V) will be

Options

A. `(V_0)/(2)`
B. `2V_(0)`
C. `V_(0) + (hc)/(2e lambda_(0))`
D. `V_(0) - (hc)/(2e lambda_(0))`

Answer

To solve the problem, we will use the principles of the photoelectric effect. The stopping potential is related to the maximum kinetic energy of the emitted electrons, which can be expressed using the equation derived from the photoelectric effect.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The kinetic energy (KE) of the emitted electrons can be expressed as:
\[ KE_{\text{max}} = E - W_0 \]
where \(E\) is the energy of the incident photons and \(W_0\) is the work function of the metal.

2. **Photon Energy**:
The energy of a photon is given by:
\[ E = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the incident light.

3. **For Wavelength \(\lambda_0\)**:
When the wavelength is \(\lambda_0\), the stopping potential \(V_0\) is given by:
\[ eV_0 = \frac{hc}{\lambda_0} - W_0 \]
Rearranging gives:
\[ eV_0 + W_0 = \frac{hc}{\lambda_0} \quad \text{(Equation 1)} \]

4. **For Wavelength \(2\lambda_0\)**:
When the wavelength is doubled to \(2\lambda_0\), the stopping potential \(V_s\) is given by:
\[ eV_s = \frac{hc}{2\lambda_0} - W_0 \]
Rearranging gives:
\[ eV_s + W_0 = \frac{hc}{2\lambda_0} \quad \text{(Equation 2)} \]

5. **Subtracting the Two Equations**:
Now, we can subtract Equation 1 from Equation 2:
\[ (eV_s + W_0) - (eV_0 + W_0) = \frac{hc}{2\lambda_0} - \frac{hc}{\lambda_0} \]
Simplifying this gives:
\[ eV_s - eV_0 = \frac{hc}{2\lambda_0} - \frac{hc}{\lambda_0} \]
\[ eV_s - eV_0 = \frac{hc}{2\lambda_0} - \frac{2hc}{2\lambda_0} = -\frac{hc}{2\lambda_0} \]

6. **Solving for Stopping Potential \(V_s\)**:
Rearranging the above equation gives:
\[ eV_s = eV_0 - \frac{hc}{2\lambda_0} \]
Dividing through by \(e\):
\[ V_s = V_0 - \frac{hc}{2e\lambda_0} \]

7. **Final Expression**:
The stopping potential when the wavelength is \(2\lambda_0\) is:
\[ V_s = V_0 - \frac{hc}{2e\lambda_0} \]