Question

Light of wavelength $\lambda_{A}$ and $\lambda_{B}$ falls on two identical metal plates $A$ and $B$ respectively. The maximum kinetic energy of photoelectrons is $K_{A}$ and $K_{B}$ respectively, then which one of the following relations is true? $\left(\lambda_{A}=2 \lambda_{B}\right)$

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

Answer 1

Answer: (A)

According to Einstein photoelectric equation,
$K_{A}=\frac{h c}{\lambda_{A}}-\phi$
$\Rightarrow 2 K_{A}=\frac{2 h c}{\lambda_{A}}-2 \phi$
or $2 K_{A}+\phi=\frac{2 h c}{\lambda_{A}}-\phi$
Given $\lambda_{A}=2 \lambda_{B}$
$\lambda_{ B }=\lambda_{ A } / 2$
$K_{B}=\frac{2 h c}{\lambda_{A}}-\phi$
$\Rightarrow 2 K_{A}+\phi=K_{B}$
$\Rightarrow K_{A}<\frac{K_{B}}{2}$