Question

Two identical metal plates show photoelectric effect. Light of wavelength ${\lambda }_A$ falls on plane A and ${\lambda }_B$ falls on plate B, ${\lambda }_A=2{\lambda }_B$. The maximum KE of the photoelectrons are $K_A$ and $K_B$, respectively. Which one of the following is true?

• A. $2K_A=K_B$
• B. $K_A=2K_B$
• C. $K_A<K_B/2$
• D. $K_A=K_B/2$

Answer 1

The correct option is C $K_A<K_B/2$

The Einstein's equation for photoelectric effect is given by $K=\frac{hc}{\lambda }-W$

where $h=$ Planck's constant , $W=$ work function for metal , $\lambda =$ wavelength of incident photon.

As metal plates are identical so work function is same for both plates i.e W (say)

For A plate : $K_A=\frac{hc}{{\lambda }_A}-W...\left(1\right)$

For B plate : $K_B=\frac{hc}{{\lambda }_B}-W...\left(2\right)$

using (1) and (2), $K_A=\frac{hc}{2{\lambda }_B}-W=\frac{1}{2}[\frac{hc}{{\lambda }_b}-2W]=\left(1/2\right)[K_B+W-2W]=K_B/2-W/2$

or $K_A+W/2=K_B/2$

Thus, $K_A<K_B/2$