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: |
$4×10^{-19}Vs$ $0.25×10^{15}Vs$ $4×10^{-15}Vs$ $4×10^{-8}Vs$ |
$4×10^{-15}Vs$ |
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 $\frac{h}{e}=\frac{λ_1λ_2(V_1-V_2)}{c(λ_2-λ_1)}=\frac{(0.6×0.4×10^{-12})(1.0)}{(3×10^8)(0.2×10^{-6})}=4×10^{-15}Vs$ |