Light of wavelength 0.6 μm from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5 V. With light of wavelength 0.4 μm from a sodium lamp, stopping potential is 1.5 V. With this data, the value of h/e is:

$4×10^{-15}Vs$

As, $eV=\frac{hc}{λ}-W$

$∴V=(\frac{h}{e}).\frac{c}{λ}-\frac{W}{e}$

$V_1=(\frac{h}{e}).\frac{c}{λ_1}-\frac{W}{e}$  (i)

$V_2=(\frac{h}{e}).\frac{c}{λ_2}-\frac{W}{e}$  (ii)

Solving these two equations, we get

$\frac{h}{e}=\frac{λ_1λ_2(V_1-V_2)}{c(λ_2-λ_1)}=\frac{(0.6×0.4×10^{-12})(1.0)}{(3×10^8)(0.2×10^{-6})}=4×10^{-15}Vs$