Question

If the wavelength of incident light changes from 400 nm to 300 nm, then the stopping potential of photoelectrons emitted from a surface becomes (approximately)

• (A) 1.0 V greater
• (B) 1.0 V smaller
• (C) 0.5 V greater
• (D) 0.5 V smaller

Answer 1

The Correct Option is A

Explanation:
As the stopping potential is given by:$V=\frac{hc}{\lambda e}-\frac{\theta }{e}$For the first case, the stopping potential is$V_1=\frac{hc}{{\lambda }_{1}e}-\frac{\varphi }{e}$For the second case, the stopping potential is$V_2=\frac{hc}{{\lambda }_{2}e}-\frac{\theta }{e}$$V_2-V_1=\frac{hc}{{\lambda }_2{e}^e}-\frac{hc}{{\lambda }_{1}e}=\frac{hc}{e}[\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}]$$V_2-V_1=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}}\times \frac{1}{{10}^{-9}}\times [\frac{1}{300}-\frac{1}{400}]$$V_2-V_1=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}}\times \frac{1}{{10}^{-9}}\times \left[\frac{1}{1200}\right]=1.03\mathrm{v}$