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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» For ground state `n_(1)=1 and n_(2)=4` Energy of photon absorbed `E=E_(2)-E_(1)` `=(-13.6)/(n_(2)^(2))-((-13.6)/(n_(1)^(2)))eV=13.6 ((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` `=13.6 ((1)/(1^(2))-(1)/(4^(2)))eV` `=13.6xx(15)/(16) eV =(13.6xx15)/(16)xx1.6xx10^(-19) J` `E=2.04xx10^(-18)` Joule. From `E=(hc)/(lamda), lamda =(hc)/(E)=(6.6xx10^(-34)xx3xx10^(-8))/(2.04xx10^(-10))=9.7xx10^(-8) m` `lamda =97 nm` `v=(c)/(lamda)=(3xx10^(8))/(9.7xx10^(-8))=3.1xx10^(15) Hz.` |
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