Question

Let $K_{1}$ and $K_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams 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. $K _{1}=\frac{ K _{2}}{3}$
• D. $K _{2}=\frac{ K _{1}}{3}$

Answer 1

Answer: (B)

$\frac{ hc }{\lambda_{1}}-\phi= K _{1}$
$\frac{ hc }{\lambda_{2}}-\phi= K _{2}$
$\lambda_{1}=3 \lambda_{2}$
$3 K _{1}=\frac{3 hc }{\lambda_{1}}-3 \phi$
$3 K _{1}=\frac{ hc }{\lambda_{2}}-3 \phi$
$3 K _{1}= K _{2}-2 \phi$
$3 K _{1}< K _{2}$
$K _{1}<\frac{ K _{2}}{3}$