Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Dual Nature of Radiation and Matter

Question:

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 ?

Options:

1 : 2

2 : 1

4 : 1

1 : 4

Correct Answer:

2 : 1

Explanation:

$W=hv_0=hc/λ_0$

$\frac{W_1}{W_2}=\frac{λ_{02}}{λ_{01}}=\frac{600}{300}=2:1$