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The fig.12.15 showes energylevel diagram of hydrogenatom . (a) Find thetransition whichresults in theemissionof a photon of wavelenght496 nm . (b) Whichtransitioncorrespondsto the emission of radiationof maximum wavelenght?Justify your answer . |
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Answer» SOLUTION :(a) Energy of photon of wavelenght `lambda = 496` NM ` = 496 xx 10^(-9) m ` is . `E = (hc)/(lambda) I = (hc)/(elambda) EV = ((6.63xx 10^(-34))xx(3xx10^(8)))/((1.6 xx 10^(-19)) xx(496 xx 10^(-9))) = 2.51 eV` From the energylevel diagram we findthat aphoton of energy2.51 eV can beemitted onlywhentransitiontakes place from n = 4ton = 2 statebecausethen `DeltaE = E_(4)- E_(2)= - 0.85 - (-3.4 eV) = + 2.55 eV` (b) Radiationof maximum wavelenghtwill be emittedphoton is minimumandit is possiblefor TRANSITION from n = 4 ton = 3 state . |
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