The photoelectric work function for a metal surface is `4.125 eV`. The cut - off wavelength for this surface is

Options

A. 4125Å
B. 3000Å
C. 6000Å
D. 2062Å

Answer

B

Explanation

Since work function for a metal surface is `W = (hc)/(lamda_0)`
where `lamda_0` is threshold wavelength or cut-off wavelength for a metal surface .
here W = 4.125 eV ` = 4.124 xx 1.6 xx 10^(-19) `Joule
so `lamda_0 = (6.6 xx 10^(-34) xx 3 xx 10^8)/(4.125 xx 1.6 xx 10^(-19)) = 3000 Å`