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)

$E=\frac{h c}{\lambda }$ or $E \propto \frac{1}{\lambda } \, \therefore \, \frac{E_{2}}{E_{1}}=\frac{\lambda _{1}}{\lambda _{2}}$
or $E_{2}=E_{1}\times \frac{\lambda _{1}}{\lambda _{2}}=1.23\times \frac{10,000}{5,000}=2.46 \, eV$
Now, $hv-\phi_{0}=\frac{1}{2}mv_{m a x}^{2}=eV_{s}$
or $\phi_{0}=hv_{2}-eV_{s}=E_{2}-eV_{s}$
$=2.46-1.36=1.10 \, eV$