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.

No, because \(\nu \)<\(\nu \)₀

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.