To determine the change in stopping potential when the wavelength of a radiation is decreased from 400 nm to 300 nm on the same metallic surface, we’ll use the concepts from the photoelectric effect. Here's the detailed, step-by-step solution:
1. Identify the given constants:
- Planck's constant, [tex]\( h = 6.62 \times 10^{-34} \, \text{Js} \)[/tex]
- Speed of light, [tex]\( c = 3 \times 10^8 \, \text{m/s} \)[/tex]
- Elementary charge, [tex]\( e = 1.6 \times 10^{-19} \, \text{C} \)[/tex]
2. Convert the wavelengths from nanometers to meters:
- Wavelength 1, [tex]\( \lambda_1 = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)[/tex]
- Wavelength 2, [tex]\( \lambda_2 = 300 \, \text{nm} = 300 \times 10^{-9} \, \text{m} \)[/tex]
3. Calculate the energy of the photons for the given wavelengths:
The energy of a photon ([tex]\( E \)[/tex]) can be calculated using the formula:
[tex]\[
E = \frac{hc}{\lambda}
\][/tex]
- Energy for [tex]\( \lambda_1 \)[/tex]:
[tex]\[
E_1 = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \, \text{J} = 4.965 \times 10^{-19} \, \text{J}
\][/tex]
- Energy for [tex]\( \lambda_2 \)[/tex]:
[tex]\[
E_2 = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} \, \text{J} = 6.62 \times 10^{-19} \, \text{J}
\][/tex]
4. Determine the change in photon energy [tex]\( \Delta E \)[/tex]:
[tex]\[
\Delta E = E_2 - E_1 = 6.62 \times 10^{-19} \, \text{J} - 4.965 \times 10^{-19} \, \text{J} = 1.655 \times 10^{-19} \, \text{J}
\][/tex]
5. Calculate the change in stopping potential [tex]\( \Delta V \)[/tex]:
Using the relation between energy and stopping potential:
[tex]\[
\Delta V = \frac{\Delta E}{e}
\][/tex]
Substituting the values we have:
[tex]\[
\Delta V = \frac{1.655 \times 10^{-19}}{1.6 \times 10^{-19}} \, \text{V} = 1.034 \, \text{V}
\][/tex]
Thus, the change in stopping potential when the wavelength of the radiation is decreased from 400 nm to 300 nm on the same metallic surface is [tex]\(1.034 \, \text{V}\)[/tex].
