Question

Let $K_{1}$ be the maximum kinetic energy of photoelectrons emitted by light of wavelength $\lambda_{1}$ and $K_{2}$ corresponding to wavelength $\lambda_{2}$. If $\lambda_{1}=2 \lambda_{2}$, then

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

Answer 1

Answer: (C)

$K_{1}< K_{2} / 2$

Answer 2

Here, $K_{1} =\frac{h c}{\lambda_{1}}-W\,\,\,\,$...(i) and $K_{2} =\frac{h c}{\lambda_{2}}-W\,\,\,\,$...(ii) Substituting $\lambda_{1}=2 \lambda_{2}$ in E (i), we get $K_{1} =\frac{h c}{2 \lambda_{2}}-W$ $K_{1} =\frac{1}{2}\left(\frac{h c}{\lambda_{2}}\right)-W$ $=\frac{1}{2}\left(K_{2}+W\right)-W$ $K_{1} =\frac{K_{2}}{2}-\frac{W}{2}$ or $ K_{1}