The wavelength of light in the visible region is about 390 nm for violet colour ,about 550 nm (average wavelength) for yellow-green colour and about 760 nm for red colour.
(a)What are the energies of photons in (eV) at the (i) violet end, (ii)average wavelength,yellow-green colour,and (iii) red end of the visible spectrum ? (Take h`=6.63xx110^(-34))` js and 1 eV=`1.6xx10^(-19)`J).
(b)From which of the photosensitive material with work functions listed n Table 11.1 and using the results of (i),(ii) and (iii) of (a), can you build a photoelectric device that operates with visible light?

Explanation

Here `lambda_(v)=390 nm=39xx10^(-8)m`
`lambda_(yg)=550 nm =55xx10^(-8)m`
`lambda_(R)=760 nm=76xx10^(-8)m`
`h=6.63xx10^(-34) Js`
leV`=1.6xx10^(-19)J`
`c=3xx10^(8)m//s`
Energy of incident photon ,
`E=hv=(hc)/(lambda)` (in joule)
`therefore E=(hc)/(elambda)to(in eV)`
(i)For violet colour,
`E_(V)=(hc)/(elambda_(V))=(6.63xx10^(-34)xx3xx10^(8))/(1.6x10^(-19)xx39xx10^(-8))`
`therefore E_(V)0.31875xx10^(1)`
`therefore E_(V)~~3.19 eV`
(ii) for yellow red light
`E_(yg)=(hc)/(elambda_(yg))=(6.63xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19)xx5xx10^(-8))`
`therefore E_(yg)=0.226xx10^(1)`
`therefore E_(eg)~~2.26 eV`
(iii) For red light,
`E_(R)=(hc)/(elambda_(R))=(6.63xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19)xx76xx10^(-8))`
`therefore E_(R)=0.163569xx10^(1)`
`therefore E_(R)~~1.64 eV`
(b)
For photoelectric emission energy of incident light should be equal or more than work function of metal.
Here energy of violet light is `E_(V)=3.19 eV` Which is more than work function of Na ,K and `C_(s)` hence photoelectric device will work which depend on radiation of violet colour.
Energy of yellow light `E_(yg)`=2.26 eV which is more than work function of `C_(S)` hence photoelectric device will work which depend on radiation of yello-green colour.
Energy of red colour is `E_(R)`=1.64 eV which is less than work function of metals given in table.Hence photoelectric device which work on red colour will not work.