The work functions for metals A,B and Care 1.92 eV, 2.0 eV and 5.0 eV respctively. The metals which will emit photoelectrons for a radiation of wavelength `4100Å` is `//` are

Options

A. A only
B. both A and B
C. all these metals
D. none

Explanation

both A and B
Energy of radiation `E= hv = (hc)/(lambda) = (((6.6 xx 10^(-34) xx 3 xx 10^(8) )// 1.6 xx 10^(-19))/(410 xx 10^(-9)))=1240/410`
E= 3.04 eV
Since energy of incident radiation is greater than the work funciton of metals A and B . So metal A and B will emit photoelectrons.