Question

Lights of wavelengths 2000 $A^0$ and 3000 $A^0$ emit photoelectrons from a metal surface whose threshold wavelength is 4000 $A^0$. The ratio of the maximum K.E’s of the photoelectrons emitted is nearly:

• A. $1:3$
• B. $3:1$
• C. $6:1$
• D. $2:1$

Answer 1

Answer: (B)

$3:1$

$Kmax=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$

$K_1=hc(\frac{1}{2000}-\frac{1}{4000})$

$K_1=\frac{hc}{4000}$

$K_2=hc(\frac{1}{3000}-\frac{1}{4000})$

$K_2=\frac{hc}{12000}$

$\frac{K_1}{K_2}=3:1$