Question

When light of wavelength 300 nm (nanometer) 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 : 2
• B. 2 : 1
• C. 4 : 1
• D. 1 : 4

Answer 1

The correct option is B 2 : 1

Work function $=\frac{hc}{{\lambda }_0};where{\lambda }_0$ is threshold wavelength.
$\therefore \frac{W_{0_1}}{W_{0_2}}=\frac{{\lambda }_{0_2}}{{\lambda }_{0_1}}=\frac{2}{1}$