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A 12.3 eV electron beam is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited ? Calculate the wavelengths of the second member of Lyman series and second member of Balmer series. |
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Answer» Solution :The energy of electron in the `N^(TH)` ORBIT ofhydrogen ATOM is `E _(n) = - ( 13.6)/( n^(2)) eV` incident beam of energy 12.3eV is absorbed by hydrogen atom. Let the electron jump fromn =1 to n =n level. `E = E_(n) -E_(1)` `12.3 = ( 13.6)/( n^(2)) - ( - ( 13.6)/1^(2))` `rArr 12.3 = 13.6 | 1- (1)/(n^(2)) | rArr ( 12.3)/( 13.6) = 1- (1)/(n^(2))` `rArr 0.9 = 1 - (1)/( n^(2)) rArr (1)/( n^(2)) = 1.09 = 0.1` `rArr n^(2) = (1)/( 0.1) = 10` `:. n = 3` That is the hydrogen atom would be excited upto second excited state. For Lyman Series `(1)/(lambda) = R[ (1)/( n_(f)^(2)) - (1)/( n_(i)^(2))]` `rArr (1)/( lambda) = 1.097 xx 10^(7) [ (1)/(1) - (1)/(9)] = 1.097 xx 10^(7) xx (8)/( 9)` ` :. lambda = ( 9)/( 1.097 xx 10^(7) xx 8) = ( 9)/( 8 xx 1.097 xx 10^(8))` `= 1.025 xx 10^(-7) = 102.5nm` For Balmer Series `(1)/( lambda) = 1.097 xx 10^(7) [ (1)/( 4) - ( 1)/( 16)]` `rArr (1)/( lambda) = 1.097 xx 10^(7) xx (3)/( 16)` `rArr lambda = 4.86 xx 10^(-7) m rArr lambda = 486nm` |
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