(a) Energy of the incident photon, E = hν = `(hc)//lambda`
`E = (6.63xx10^(–34)J s) (3xx10^(8)
m//s)//lambda`
`=(1.989xx10^(25) J m )/lambda`
(i)For violet light, `lambda_1=390 nm` (lower wavelength end )
Incident photon energy , `E_1=(1.989xx10^(-25) Jm)/(390xx10^(-9) m)`
`= 5.10 xx 10^(–19)J`
`=( 5.10 xx 10^(–19)J)/(1.6xx10^(-19) J//eV)`
=3.19 eV
(ii)For yellow-green light, `lambda_2=550 nm` (average wavelength )
Incident photon energy , `E_2=(1.989xx10^(-25) Jm)/(550xx10^(-9)m)`
`=3.62xx10^(-19) J =2.26 eV`
(iii) For red light, `lambda_3` = 760 nm (higher wavelength end)
Incident photon energy, `E_3=(1.989xx10^(-25) Jm)/(760xx10^(-9)m)`
`=2.62xx10^(-19) J =1.64 eV`
(b) For a photoelectric device to operate, we require incident light energy E to be equal to or greater than the work function `phi_0` of the material. Thus, the photoelectric device will operate with violet light (with E = 3.19 eV) photosensitive material Na (with `phi_0` = 2.75 eV), K (with `phi_0` = 2.30 eV) and Cs (with `phi_0` = 2.14 eV). It will also operate with yellow-green light (with E = 2.26 eV) for Cs (with `phi_0` = 2.14 eV) only. However, it will not operate with red light (with E = 1.64 eV) for any of these photosensitive materials.