The wavelength of light in the visible region is about 390nm for violet color, about 550nm (average wavelength) for yellow green colour and about 760n
The wavelength of light in the visible region is about 390nm for violet color, about 550nm (average wavelength) for yellow green colour and about 760nm for red color. What are the energies of photon in eV at the (i) violet end (ii) average wavelength yellow-green color and (iii) red end of the visible spectrum? `("Take " h=6.63xx10^(-34)Js and 1eV=1.6xx10^(-19)Js)`
1 Answers
(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.