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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? |
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Answer» Solution :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. |
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