Question 241:
easyWhen 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?
(1993)
The work function is inversely proportional to the threshold wavelength: $\phi = \frac{hc}{\lambda_0}$. The ratio of the work functions is $\frac{\phi_1}{\phi_2} = \frac{\lambda_{02}}{\lambda_{01}} = \frac{600 nm}{300 nm} = \frac{2}{1}$. Thus, the ratio is 2 : 1.