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. $K_1>\frac{K_2}{3}$
• B. $K_1<\frac{K_2}{3}$
• C. $K_1=3K_2$
• D. $K_2=3K_1$

Answer 1

The correct option is B $K_1<\frac{K_2}{3}$

According to Einstein's photoelectric

$\frac{hc}{\lambda }={\varphi }_0+K{E}_{max}$

$\therefore K_1=\frac{hc}{{\lambda }_1}-{\varphi }_0$ and $K_2=\frac{hc}{{\lambda }_2}-{\varphi }_0$

or $K_1-K_2=hc[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}]$

=$hc[\frac{1}{3{\lambda }_2}-\frac{1}{{\lambda }_2}]=-\frac{2hc}{3{\lambda }_2}$ (Given ${\lambda }_1=3{\lambda }_2)$

$K_1-K_2=-\frac{2}{3}(K_2+{\varphi }_0)$

or $K_1=K_2-\frac{2}{3}{K}_2-\frac{2}{3}{\varphi }_0=\frac{K_2}{3}-\frac{2}{3}{\varphi }_0$

or $K_1<\frac{K_2}{3}$