Question

Two identical metal plates show the photoelectric effect by a light of wavelength $\lambda _{1}$ on plate 1 and $\lambda _{2}$ on plate 2 (where $\lambda _{1}=2\lambda _{2}$ ). Then the relation between kinetic energies will be

• A. $2K_{2}=K_{1}$
• B. $K_{1} < \frac{K_{2}}{2}$
• C. $K_{1}>\frac{K_{2}}{2}$
• D. $2K_{1}=K_{2}$

Answer 1

Answer: (B)

According to Einstein photelectric equation
$K.E=\frac{h c}{\lambda }-\phi$
$\frac{h c}{\lambda _{1}}=KE_{1}+\phi$ .... (i) (for $\lambda _{1}$
$\frac{h c}{\lambda _{2}}=KE_{2}+\phi$ .... (ii) (for $\lambda _{2}$
Let $\lambda _{1}=\lambda \, $ and $\lambda _{2}=\frac{\lambda _{1}}{2}=\frac{\lambda }{2}$
From Eqs. (i) and (ii),
$\frac{K E_{2}}{K E_{1}}=\frac{\frac{h c}{\lambda _{2}} - \phi}{\frac{h c}{\lambda _{1}} - \phi}=\frac{\frac{h c}{\frac{\lambda }{2}} - \phi}{\frac{h c}{\lambda } - \phi}=\frac{\frac{2 h c}{\lambda } - \phi}{\frac{h c}{\lambda } - \phi}$
$=\frac{\frac{h c}{\lambda } - \phi}{\frac{h c}{\lambda } - \phi}+\frac{\frac{h c}{\lambda }}{\frac{h c}{\lambda } - \phi}$
$\because \frac{\frac{h c}{\lambda }}{\frac{h c}{\lambda } - \phi}>1$
$\therefore \frac{K E_{2}}{K E_{1}}=1+\left(> 1\right)>2$
Thus, $KE_{1} < \frac{K E_{2}}{2}$