Question

$K_1$and $K_2$ are the maximum kinetic energies of the photoelectrons emitted when light of wavelength ${\lambda }_1$ and ${\lambda }_2$ respectively are incident on a metallic surface. If ${\lambda }_1=$3${\lambda }_2$ then

• A. $K_1>\frac{K_2}{3}$
• B. $K_1<\frac{K_2}{3}$
• C. $K_1=3K_2$
• D. $K_2=3K_1$

Answer 1

Answer: (B)

$K_1<\frac{K_2}{3}$

$K_1=\frac{hc}{{\lambda }_1}-\varphi$-----------(1)

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

$\because {\lambda }_1=3{\lambda }_2$

$k_2=3\left(\frac{hc}{{\lambda }_1}\right)-\varphi$---------(3)

$\frac{k_2}{3}=\frac{hc}{{\lambda }_1}-\frac{\varphi }{3}$----------(4)

$\frac{k_2}{3}=(k_1+\varphi )-\frac{\varphi }{3}$ (by (1))

$\frac{k_2}{3}=k_1+\frac{2\varphi }{3}$

So,$k_1<\frac{k_2}{3}$