When a surface is irradiated with light of λ = 4950 Å, a photocurrent appears which vanishes if a retarding potential greater than 0.6 V is applied across the photo tube. When a different source of light is used, it is found that the critical retarding potential is changed to 1.1 V. What is the work function of the surface and the wavelength of the second source? If the photoelectrons (after emission from the source) are subjected to a magnetic field of 10 tesla what changes will be observed in the above two retarding potentials?

where,
ɸ₀ is the work function,
X wavelength of incident light and
V₀ is the stopping potential. 

For the first source,
                           $$\begin{gathered} {\mathrm{\lambda }}_1 = 4950 \mathrm{\AA } = 4950 \times {10}^{-10}\mathrm{m} \\ {\mathrm{V}}_0 = 0.6 \mathrm{V} \end{gathered}$$ 

Putting these values in equation 1, 
$\therefore 1.6 \times {10}^{-19} \times 0.6 = \frac{6.6 \times {10}^{-34} \times 3 \times {10}^8}{495 \times {10}^{-9}}-{\mathrm{\varphi }}_0$

$\Rightarrow$             $0.96\times {10}^{-19} = 4\times {10}^{-19}-{\mathrm{\varphi }}_0$ 

$\therefore$                             ${\mathrm{\varphi }}_0 = 3.04 \times {10}^{-19}\mathrm{J}$             ...(I)
                                     $$\begin{gathered} = \frac{3.04 \times {10}^{-19}}{1.6 \times {10}^{-19}}\mathrm{eV} \\ = 1.9 \mathrm{eV} \end{gathered}$$ 

Let λ₂ be the wavelength of the second source.
Stopping potential, V₀’ = 1.1 V (given) 

Therefore, $1.6 \times {10}^{-19} \times 1.1 = \frac{6.6 \times {10}^{-34}\times 3\times {10}^8}{{\mathrm{\lambda }}_2}-3.04 \times {10}^{-19}\mathrm{J} (\mathrm{from} \mathrm{I})$
$\Rightarrow$  $1.76 \times {10}^{-19} = \frac{19.8 \times {10}^{-26}}{{\mathrm{\lambda }}_2}-3.04 \times {10}^{-19}$ 

$\Rightarrow$          $\frac{19.8 \times {10}^{-26}}{{\mathrm{\lambda }}_2} = 4.8 \times {10}^{-19}$
$\therefore$    ${\mathrm{\lambda }}_2 = \frac{19.8 \times {10}^{-26}}{4.8 \times {10}^{-19}}\mathrm{m} = 4.125 \times {10}^{-7}\mathrm{m}$
                                          $= 4125 \mathrm{\AA }$ 

When the ejected photoelectrons are subjected to the action of a magnetic field no change in retarding potential will be observed because, a magnetic field does not alter the kinetic energy of the photoelectrons. The magnetic field only changes the direction of motion.