Question

When light of wave length 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, however, light of 600nm wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters?

• A. 1:2
• B. 2:1
• C. 1:4
• D. 4:1

Answer 1

The correct option is B 2:1

Work function $\varphi =\frac{hc}{\lambda }$
$⟹$ $\frac{{\varphi }_1}{{\varphi }_2}=\frac{{\lambda }_2}{{\lambda }_1}$
Given : ${\lambda }_1=300nm$ ${\lambda }_2=600nm$
$\therefore$ $\frac{{\varphi }_1}{{\varphi }_2}=\frac{600}{300}=2$