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
$K.E=\frac{hc}{\lambda }-\varphi$
$\frac{hc}{{\lambda }_1}=K{E}_1+\varphi$ .... (i) (for ${\lambda }_1$
$\frac{hc}{{\lambda }_2}=K{E}_2+\varphi$ .... (ii) (for ${\lambda }_2$
Let ${\lambda }_1=\lambda$ and ${\lambda }_2=\frac{{\lambda }_1}{2}=\frac{\lambda }{2}$
From Eqs. (i) and (ii),
$\frac{K{E}_2}{K{E}_1}=\frac{\frac{hc}{{\lambda }_2}-\varphi }{\frac{hc}{{\lambda }_1}-\varphi }=\frac{\frac{hc}{\frac{\lambda }{2}}-\varphi }{\frac{hc}{\lambda }-\varphi }=\frac{\frac{2hc}{\lambda }-\varphi }{\frac{hc}{\lambda }-\varphi }$
$=\frac{\frac{hc}{\lambda }-\varphi }{\frac{hc}{\lambda }-\varphi }+\frac{\frac{hc}{\lambda }}{\frac{hc}{\lambda }-\varphi }$
$\because \frac{\frac{hc}{\lambda }}{\frac{hc}{\lambda }-\varphi }>1$
$\therefore \frac{K{E}_2}{K{E}_1}=1+\left(>1\right)>2$
Thus, $K{E}_1<\frac{K{E}_2}{2}$