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A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n=4 level. Determine the wavelength and frequency of photon. |
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Answer» Solution : As initially the hydrogen atom is in its ground STATE n = 1, its initial energy `E_(i)=- (13.6)/((1)^(2))= - 13.6 eV` . On excitation to n = 4 LEVEL, new energy of electron becomes `E_(f) = (13.6)/((4)^(2))= -0.85 eV` . `therefore ` Energyof photon absorbed `E = E_(f) - E_(i) = (-0.85) - (-13.6) eV` ` = 12.75 eV = 12.75 xx 1.6 xx 10^(-9)J` `therefore ` Frequencyof absorbed photo `v = (E)/(H) = (12.75 xx 1.6 xx 10^(-19))/(6.63 xx 10^(-34)) = 3.1 xx 10^(15 )HZ` and`""` wavelenght`lambda = (c)/(v) = (3 xx 10^(8))/(3.1 xx 10^(15)) = 9.7 xx 10^(-8) m` |
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