In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. Choose the closest value of change in the stopping potential from given options `((hc)/(e)=1240 nm.V)`

Options

A. 2.0 V
B. 0.5 V
C. 1.0 V
D. 1.5 V

Answer

To solve the problem, we need to determine the change in stopping potential when the wavelength of the incident light changes from 300 nm to 400 nm in a photoelectric experiment. We will use the photoelectric equation and the relationship between stopping potential and wavelength.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The photoelectric effect can be described by the equation:
\[ E = \phi + eV \]
where \( E \) is the energy of the incident photons, \( \phi \) is the work function of the metal, \( e \) is the charge of the electron, and \( V \) is the stopping potential.

2. **Photon Energy Calculation**:
The energy of a photon can be expressed in terms of its wavelength:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light. Given that \( \frac{hc}{e} = 1240 \, \text{nm.V} \), we can express the energy in terms of stopping potential.

3. **Set Up the Equations**:
For the two wavelengths, we can write:
- For \( \lambda_1 = 300 \, \text{nm} \):
\[ \frac{hc}{\lambda_1} = \phi + eV_1 \]
- For \( \lambda_2 = 400 \, \text{nm} \):
\[ \frac{hc}{\lambda_2} = \phi + eV_2 \]

4. **Subtract the Two Equations**:
Subtract the second equation from the first:
\[ \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} = eV_1 - eV_2 \]
This simplifies to:
\[ e(V_1 - V_2) = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]

5. **Express Change in Stopping Potential**:
Rearranging gives:
\[ V_1 - V_2 = \frac{hc}{e} \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]
Substituting \( \frac{hc}{e} = 1240 \, \text{nm.V} \):
\[ V_1 - V_2 = 1240 \left( \frac{1}{300} - \frac{1}{400} \right) \]

6. **Calculate the Change**:
First, calculate \( \frac{1}{300} - \frac{1}{400} \):
\[ \frac{1}{300} = \frac{4}{1200}, \quad \frac{1}{400} = \frac{3}{1200} \]
Thus,
\[ \frac{1}{300} - \frac{1}{400} = \frac{4 - 3}{1200} = \frac{1}{1200} \]
Now substitute this back:
\[ V_1 - V_2 = 1240 \times \frac{1}{1200} = \frac{1240}{1200} \approx 1.033 \, \text{V} \]

7. **Final Answer**:
The closest value of change in stopping potential is approximately 1 V.