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Question

Physics Question on Photoelectric Effect

When light of wavelength 300 nm falls on a photoelectric 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 functions of the two emitters:

A

1:04

B

4:01

C

2:01

D

1:02

Answer

2:01

Explanation

Solution

From the formula of work function W0=hcλ{{W}_{0}}=\frac{hc}{\lambda } here λ0{{\lambda }_{0}} is threshold wavelength W01W02=λ02λ01=600300=21\frac{{{W}_{01}}}{{{W}_{02}}}=\frac{{{\lambda }_{02}}}{{{\lambda }_{01}}}=\frac{600}{300}=\frac{2}{1}