Question

A photo sensitive metal is incident with radiations of wavelength 400 nm and then with radiations of wavelength 800 nm. What will be the difference in the maximum energy of the photoelectrons ?

• A. 0.5 eV
• B. 1.0 eV
• C. 1.5 eV
• D. 2 eV

Answer 1

Answer: (C)

$K_1=\frac{hc}{{\lambda }_1}-{\omega }_0$ $K_2=\frac{hc}{{\lambda }_2}-{\omega }_0$ $K_1-K_2=hc\left[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right]=hc\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_2{\lambda }_1}$ = $\frac{6.6\times 10^{-34}\times 3\times 10^8[800-400]\times 10^{-9}}{400\times 800\times 10^{-18}}=\frac{2.475\times 10^{-19}}{J}=$ 1.5 eV