Question

Light of wavelength ${\lambda }_A$ and ${\lambda }_B$ falls on two identical metal plates A and B respectively. The maximum kinetic energy of photo-electrons in $K_A$ and $K_B$ respectively, then which one of the following relations is true? $({\lambda }_A=2{\lambda }_B)$

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

Answer 1

The correct option is A $K_A<\frac{K_B}{2}$

Let $T$ denote the work function of the material.

Since the plates are identical, $T=T_A=T_B$.

Work function is the minimum energy of the incident beam necessary to pull out an electron from the metal surface. Any excess energy of the radiation manifests itself as the kinetic energy of the ejected electron. Energy of incident photon of wavelength $\lambda$ is $\frac{hc}{\lambda }$.

So, we have:

$K_A=\frac{hc}{{\lambda }_A}-T{K}_B=\frac{hc}{{\lambda }_B}-T=\frac{2hc}{{\lambda }_A}-T>2(\frac{hc}{{\lambda }_A}-T)=2K_A\Rightarrow K_A<\frac{K_B}{2}$