Question

Two identical metal plates show photoelectric effect by a light of wavelength ${\lambda }_1$ on plate 1 and ${\lambda }_2$ on plate 2 (where ${\lambda }_1=2{\lambda }_2$ ). The maximum kinetic energy will be :

• A. $2K_2=K_1$
• B. $K_1<K_2/2$
• C. $K_1>K_2/2$
• D. $2K-1=K_2$

Answer 1

Answer: (B)

$K_1<K_2/2$

Maximum kinetic energy $K=\frac{hc}{\lambda }-\varphi$

where $\lambda$ is the wavelength of incident light and $\varphi$ is the work function of the metal.

For first case : $\lambda ={\lambda }_1=2{\lambda }_2$

$K_1=\frac{hc}{2{\lambda }_2}-\varphi$ ....(1)

For second case : $\lambda ={\lambda }_2$

$K_2=\frac{hc}{{\lambda }_2}-\varphi$

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

From (2) - (1), we get

$\frac{K_2}{2}-K_1=\frac{\varphi }{2}$ $⟹\frac{K_2}{2}>K_1$