To solve the problem of finding the change in stopping potential when the wavelength of light incident on a photosensitive material is changed, we can use the photoelectric equation:
\[
K.E. = h \nu - \phi
\]
where \( K.E. \) is the kinetic energy of the photoelectrons, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the material.
The stopping potential \( V_0 \) is given by:
\[
K.E. = eV_0
\]
The frequency \( \nu \) of the light can be related to its wavelength \( \lambda \) using the equation:
\[
\nu = \frac{c}{\lambda}
\]
Where \( c \) is the speed of light. Substituting this into the photoelectric equation gives:
\[
eV_0 = h \left(\frac{c}{\lambda}\right) - \phi
\]
The change in stopping potential when the wavelength changes from \( \lambda_1 = 2000 Å \) to \( \lambda_2 = 2100 Å \) can be calculated as follows:
\[
\Delta V_0 = V_0(\lambda_2) - V_0(\lambda_1)
\]
Expressing \( V_0 \) in terms of wavelengths:
\[
e\Delta V_0 = h \left(\frac{c}{\lambda_2} - \frac{c}{\lambda_1}\right)
\]
Rearranging gives:
\[
\Delta V_0 = \frac{hc}{e} \left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)
\]
Assuming the known constants:
- Planck's constant \( h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}\)
- Speed of light \( c = 3 \times 10^{8} \text{ m/s}\)
- Electron charge \( e = 1.6 \times 10^{-19} \text{ C}\)
Converting the wavelengths from angstroms to meters:
- \( \lambda_1 = 2000 \text{ Å} = 2000 \times 10^{-10} \text{ m}\)
- \( \lambda_2 = 2100 \text{ Å} = 2100 \times 10^{-10} \text{ m}\)
Substitute these values into the equation:
\[
\Delta V_0 = \left(\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19}}\right) \left(\frac{1}{2100 \times 10^{-10}} - \frac{1}{2000 \times 10^{-10}}\right)
\]
\[
\Delta V_0 = \left(1.239 \times 10^{-6}\right) \left(\frac{1}{2.1 \times 10^{-7}} - \frac{1}{2.0 \times 10^{-7}}\right)
\]
Calculate the difference in frquency terms:
\[
\Delta V_0 = (1.239 \times 10^{-6}) \left(\frac{0.1}{4.2} \times 10^{7}\right)
\]
\[
\Delta V_0 = 1.239 \times 10^{-6} \times 4.76 \times 10^{6}
\]
\[
\Delta V_0 = 0.3 V
\]
Therefore, the change in stopping potential is 0.3 V, which confirms the correct answer as 0.3 V.