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:

1.10 eV

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 $\phi_0=hv_2-eV_s=E_2-eV_s=2.46-1.36=1.10eV$