In a photoelectric effect experiment, the stopping potential is 0.71 V and corresponding wavelength of incident photon was 491 nm. Now the stopping potential of battery is increased to 1.4 V. Find new wavelength of incident photon ?

Options

A. `390 nm`
B. `321 nm`
C. `275 nm`
D. `392 nm`

Answer

To solve the problem, we will use the concept of the photoelectric effect, which relates the energy of the incident photons to the stopping potential and the work function of the material.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The energy of the incident photon can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)), \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)), and \( \lambda \) is the wavelength of the incident light.

2. **Relating Energy to Stopping Potential**:
The stopping potential \( V_0 \) is related to the kinetic energy of the emitted electrons:
\[ E = eV_0 + \phi \]
where \( e \) is the elementary charge (\( 1.6 \times 10^{-19} \, \text{C} \)) and \( \phi \) is the work function of the material.

3. **Setting Up the Equations**:
For the first condition (with \( V_0 = 0.71 \, \text{V} \) and \( \lambda = 491 \, \text{nm} \)):
\[ \frac{hc}{491 \times 10^{-9}} = e \cdot 0.71 + \phi \]
For the second condition (with \( V_0 = 1.4 \, \text{V} \) and unknown \( \lambda_2 \)):
\[ \frac{hc}{\lambda_2} = e \cdot 1.4 + \phi \]

4. **Substituting Values**:
We can substitute \( h \) and \( c \):
\[ hc = 1240 \, \text{eV nm} \]
Thus, the first equation becomes:
\[ \frac{1240}{491} = 0.71 + \frac{\phi}{e} \]
And the second equation becomes:
\[ \frac{1240}{\lambda_2} = 1.4 + \frac{\phi}{e} \]

5. **Subtracting the Equations**:
Subtract the first equation from the second:
\[ \frac{1240}{\lambda_2} - \frac{1240}{491} = 1.4 - 0.71 \]
Simplifying gives:
\[ \frac{1240}{\lambda_2} - \frac{1240}{491} = 0.69 \]

6. **Finding \( \lambda_2 \)**:
Rearranging gives:
\[ \frac{1240}{\lambda_2} = 0.69 + \frac{1240}{491} \]
Calculate \( \frac{1240}{491} \):
\[ \frac{1240}{491} \approx 2.53 \]
Thus:
\[ \frac{1240}{\lambda_2} = 0.69 + 2.53 = 3.22 \]
Therefore:
\[ \lambda_2 = \frac{1240}{3.22} \approx 385.09 \, \text{nm} \]

7. **Final Answer**:
The new wavelength of the incident photon is approximately:
\[ \lambda_2 \approx 385 \, \text{nm} \]