Question

The stopping potentials are V₁ and V₂ with incident lights of wavelength l₁ and l₂ respectively. Then V₁ – V₂-

• A. $\frac{hc}{e}\left( \frac{\lambda_{1}\lambda_{2}}{\lambda_{1}–\lambda_{2}} \right)$
• B. $\frac{hc}{e}\left( \frac{1}{\lambda_{1}}–\frac{1}{\lambda_{2}} \right)$
• C. $\frac{he}{c}\left( \frac{1}{\lambda_{1}}–\frac{1}{\lambda_{2}} \right)$
• D. $\frac{he}{c\lambda_{1}\lambda_{2}}\left( \lambda_{1}–\lambda_{2} \right)$

Answer 1

Answer: (B)

$\frac{hc}{e}\left( \frac{1}{\lambda_{1}}–\frac{1}{\lambda_{2}} \right)$

Answer 2

eV₁ =$\frac { \mathrm { hc } } { \lambda _ { 1 } }$– f

eV₂ = – f

V₁ – V₂ = $\frac { \mathrm { hc } } { \mathrm { e } }$ $\left( \frac { 1 } { \lambda _ { 1 } } - \frac { 1 } { \lambda _ { 2 } } \right)$