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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 about760 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 (ii) red end of the visible spectrum ? (Take h = 6.63 xx 10^(-34) Js and 1 eV=1.6 xx 10^(-19) J ) , |
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Answer» Solution :(a) Eneergy of the incident photon, `E=hv=hc//lambda` `E=(6.63xx10^(-34)Js)(3xx10^(8)m//s)//lambda` `=(1.989xx10^(-25)Jm)/(lambda)` (i) For violet LIGHT, `lambda_(1)=390nm" (lower wavelength end)"` `"Incident photon energy, "E_(1)=(1.989xx10^(-25)Jm)/(390xx10^(-9)m)` `=5.10xx10^(-19)J` `=(5.10xx10^(-19)J)/(1.6xx10^(-19)J//eV)` `=3.19eV` (II) For YELLOW - green light, `lambda_(2)=550nm"(average wavelength)"` `"Incident photon energy, "E_(2)=(1.989xx10^(-25)Jm)/(550xx10^(-9)m)` `=3.62xx10^(-19)J=2.26ev` (iii) For red light, `lambda_(3)=760nm"(higher wavelength end)"` `"Incident photon energy, "E_(3)=(1.989xx10^(-25)Jm)/(760xx10^(-9)m)` `=2.62xx10^(-19)J=1.64ev` (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. |
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