Question

If $K_1$ and $K_2$ are maximum kinetic energies of photoelectrons emitted when lights of wavelengths ${\lambda }_1$ and ${\lambda }_2$ respectively are incident on a metallic surface. If ${\lambda }_1=3{\lambda }_2$, then:

• A. $K_1>\left(K_2/3\right)$
• B. $K_1<\left(K_2/3\right)$
• C. $K_1=2K_2$
• D. $K_2=2K_1$

Answer 1

The correct option is B $K_1<\left(K_2/3\right)$

Kinetic energy of the photoelectrons $K=\frac{hc}{\lambda }-\varphi$ where $\varphi$ is the work function of the metal

$\therefore$ For wavelength ${\lambda }_1$, $K_1=\frac{hc}{{\lambda }_1}-\varphi$ ............(1)

For wavelength ${\lambda }_2$, $K_2=\frac{hc}{{\lambda }_2}-\varphi$ .........(2)

Given : ${\lambda }_1=3{\lambda }_2$

$\therefore$ Equation (1) becomes $K_1=\frac{hc}{3{\lambda }_2}-\varphi$ ............(3)

From (2) - (3), we get $K_2-K_1=$ $\frac{hc}{{\lambda }_2}-\frac{hc}{3{\lambda }_2}$

$K_2-K_1=$ $\frac{2}{3}\frac{hc}{{\lambda }_2}$ $⟹\frac{hc}{{\lambda }_2}=\frac{3}{2}(K_2-K_1)$

Put this in (2), $K_2=\frac{3}{2}(K_2-K_1)-\varphi$

$⟹$ $K_2-3K_1=2\varphi$

As $\varphi >0$ $⟹K_2-3K_1>0$

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