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
$\frac{hc}{\lambda }={\varphi }_0+K{E}_{max}$
$\therefore K_1=\frac{hc}{{\lambda }_1}-{\varphi }_0$ and $K_2=\frac{hc}{{\lambda }_2}-{\varphi }_0$
or $K_1-K_2=hc\left[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right]$
$=hc\left[\frac{1}{3{\lambda }_2}-\frac{1}{{\lambda }_2}\right]=-\frac{2hc}{3{\lambda }_2}$ (Given ${\lambda }_1=3{\lambda }_2)$
$K_1-K_2=-\frac{2}{3}\left(K_2+{\varphi }_0\right)$
or $K_1=K_2-\frac{2}{3}{\varphi }_0=\frac{K_2}{3}-\frac{2}{3}{\varphi }_0$ or $K_1<\frac{K_2}{3}$