Question

Given that a photon of light of wavelength 10,000 $Å$ has an energy equal to 1.23 eV. When light of wavelength 5000 $Å$ and intensity $I_0$ falls on a photoelectric cell, the surface current is $0.40\times 10^{-6}$ A and the stopping potential is 1.36 V, then the work function is

• A. 0.43 eV
• B. 0.55 eV
• C. 1.10 eV
• D. 1.53 eV

Answer 1

Answer: (C)

$E=\frac{hc}{\lambda }$ or $E\propto \frac{1}{\lambda }\therefore \frac{E_2}{E_1}=\frac{{\lambda }_1}{{\lambda }_2}$
or $E_2=E_1\times \frac{{\lambda }_1}{{\lambda }_2}=1.23\times \frac{10,000}{5,000}=2.46eV$
Now, $hv-{\varphi }_0=\frac{1}{2}m{v}_{max}^2=e{V}_s$
or ${\varphi }_0=h{v}_2-e{V}_s=E_2-e{V}_s$
$=2.46-1.36=1.10eV$