A radiation of wavelength 300 nm is incident on a silver surface. Will photoelectrons be observed ?

Explanation

Energy of the incident photon is ,
`E = hv = (hc)/(lambda)` ( in joules )
`E= (hc)/(lambda_e)` ( in eV)
Substituting the known values , we get
`E=(6.626 xx 10^(-34) xx 3 xx 10^(8))/(300 xx 10^(-9) xx 1.6 xx 10^(-19))`
E=4.14 eV
The work function of silver = 4.7 eV . Since the energy of the incident photon is less than the work function of silver , photoelectrons are not observed in this case .