Given that a photon of light of wavelength 10,000° has an energy equal to 1.23 eV. When light of wavelength 5000Å and intensity I0 falls on a photoelectric cell, the surface current is $0.40×10^{−6} A$ and the stopping potential is 1.36 V, then the work function is:
Answer & explanation
Correct answer: option 3
Energy of a photon $E=\frac{hc}{λ}$
or $E∝\frac{1}{λ}$ $∴\frac{E_2}{E_1}=\frac{λ_1}{λ_2}$
or $E_2=E_1×\frac{λ_1}{λ_2}=1.23×\frac{10000}{5000}=2.46eV$
According to Einstein’s photoelectric equation
$hv-\phi_0=\frac{1}{2}mv^2_{max}=eV_s$
or $\phi_0=hv_2-eV_s=E_2-eV_s=2.46-1.36=1.10eV$