Question

If $K_{1}$ and $K_{2}$ are maximum kinetic energies of photoelectrons emitted when lights of wavelength $\lambda_{1}$ and $\lambda_{2}$ respectively incident on a metallic surface if $\lambda_{1}=3\lambda_{2}$, then

• A. $K_{1} > \, (K_{2}/3)$
• B. $K_{1}<\, (K_{2}/3)$
• C. $K_{1}=3K_{2}$
• D. $K_{2}=3K_{1}$

Answer 1

Answer: (B)

According to Einsteins photoelectric equation
$\frac{hc}{\lambda}=\phi_{0}+KE_{max}$
$\therefore K_{1}=\frac{hc}{\lambda_{1}}-\phi_{0}$ and $K_{2}=\frac{hc}{\lambda_{2}}-\phi_{0}$
or $K_{1}-K_{2}=hc \left[\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right]$
$=hc\left[\frac{1}{3\lambda_{2}}-\frac{1}{\lambda_{2}}\right]=-\frac{2hc}{3\,\lambda_{2}}$ (Given $\lambda_{1}=3\lambda_{2})$
$K_{1}-K_{2}=-\frac{2}{3}\left(K_{2}+\phi_{0}\right)$
or $K_{1}=K_{2}-\frac{2}{3} \phi_{0} =\frac{K_{2}}{3}-\frac{2}{3} \phi_{0}$ or $K_{1}<\, \frac{K_{2}}{3}$