Question

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

• A. 1:2
• B. 2:1
• C. 4:1
• D. 1:4

Answer 1

The correct option is B 2:1

$\frac{hc}{\lambda }=\varphi \Rightarrow \varphi \alpha \frac{1}{\lambda };\frac{{\varphi }_1}{{\varphi }_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{600}{300}$ = 2 : 1