A specific metal's work function is 4.2 eV. Will this metal emit photoelectrically for incident light with a 330 nm wavelength? Also mark the correct reason for the answer.
Answer & explanation
Correct answer: option 2
The minimum threshold frequency is given by
\(\nu \)0 = \(\phi\)0/h
metal's work function \(\phi\)0 = 4.2 eV
h = 6.6×10-34 Js
So, \(\nu \)0 = 4.2×1.6×10-19/6.6×10-34
\(\nu \)0 = 1.018×1015 Hz ...............................(1)
the relation between frequency and wavelength is given by
\(\nu \) = c/\(\lambda \)
= 3 × 108/330 × 10-9
=0.9 ×1015 Hz .................................(2)
So from ...(1) and ...(2) it is clear that \(\nu \)<\(\nu \)0, so this metal will NOT emit photoelectrically.