If the wavelength of incident light changes from 400 nm to 300 nm, the stopping potential for photoelectrons emitted from a surface becomes approximately

Options

A. 1.0 V greater
B. 1.0 V smaller
C. 0.5 V greater
D. 0.5 V smaller

Explanation

To solve the problem, we need to understand the relationship between the wavelength of incident light and the stopping potential for photoelectrons emitted from a surface. The stopping potential (V₀) is related to the energy of the incident photons, which can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the light.

The stopping potential is also related to the energy of the emitted photoelectrons by the equation:

\[ eV_0 = E \]

where:
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \)),
- \( V_0 \) is the stopping potential.

### Step-by-Step Solution:

1. **Calculate the energy of the incident photons for both wavelengths.**

For \( \lambda_1 = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \):

\[ E_1 = \frac{hc}{\lambda_1} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \]

\[ E_1 = \frac{1.9878 \times 10^{-25}}{400 \times 10^{-9}} = 4.9695 \times 10^{-19} \, \text{J} \]

For \( \lambda_2 = 300 \, \text{nm} = 300 \times 10^{-9} \, \text{m} \):

\[ E_2 = \frac{hc}{\lambda_2} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{300 \times 10^{-9} \, \text{m}} \]

\[ E_2 = \frac{1.9878 \times 10^{-25}}{300 \times 10^{-9}} = 6.626 \times 10^{-19} \, \text{J} \]

2. **Calculate the stopping potential for both cases.**

Using the relation \( eV_0 = E \):

For \( E_1 \):

\[ V_{0_1} = \frac{E_1}{e} = \frac{4.9695 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.105 \, \text{V} \]

For \( E_2 \):

\[ V_{0_2} = \frac{E_2}{e} = \frac{6.626 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 4.141 \, \text{V} \]

3. **Determine the change in stopping potential.**

The change in stopping potential when the wavelength changes from 400 nm to 300 nm is:

\[ \Delta V_0 = V_{0_2} - V_{0_1} = 4.141 \, \text{V} - 3.105 \, \text{V} \approx 1.036 \, \text{V} \]

### Final Answer:

The stopping potential for photoelectrons emitted from the surface approximately becomes \( 4.141 \, \text{V} \) when the wavelength changes from 400 nm to 300 nm.