Q. When light of wavelength (nanometer) falls on a photoelectric emitter, photoelectrons are just liberated. For another emitter, however, light of wavelength is sufficient for creating photoemission. What is the ratio of the work function of the two emitters?

 2448  225 J & K CETJ & K CET 2003 Report Error

Solution:

The minimum energy required for the emission of photoelectrons from a metal is called the work function of that metal.
Given by (i)
where is threshold frequency and is Planck's constant.
Also, ...(ii)
where is speed of light and the threshold wavelength.
From Eqs. (i) and (ii), we get

Given,