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  • When light of wavelength (nanometer) falls on a photoelectric emitter, photoelectrons are liberated. For another

When light of wavelength 300 nm (nanometer) 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 ?

Option: 1

1 : 2


Option: 2

2 : 1


Option: 3

4 : 1


Option: 4

1 : 4


Answers (1)

best_answer

\begin{aligned} & \mathrm{W}=\mathrm{h} v_0=\mathrm{hc} / \lambda_0 \\ & \frac{\mathrm{W}_1}{\mathrm{~W}_2}=\frac{\lambda_{02}}{\lambda_{01}}=\frac{600}{300}=2: 1 \end{aligned}

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Ritika Harsh

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