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The IP_(1) of His 13.6 eV it is expoxedinduced radiation .Find the wavelengthof these ijnduced radiation |
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Answer» Solution :`E_(1)` of H atom `= - 13.6 eV`1 Energy givento H atom `= (6.625xx 10^(-34) xx 3.0 xx 10^(8))/(1028 xx 10^(-10))` `= 1.933 xx 10^(-18) J = 12.07 eV` Energy of H atom after excitation `= - 13.6 + 1207 = - 1.53 eV` `:. E_(N) = (E_(1))/(n^(2))` `:. n^(2) = (-13.6)/(-1.53) = 9 :. n = 3` The electron in H atom is excited to third shell :. induced `lambda_(1)= (HC)/(E_(3) - E_(1))` `E_(1) = -13.6 ev, E_(3) = - 1.53 eV` `lambda_(1) = (6.626 xx 110^(-34) xx 3 xx 10^(6))/((-1.53 + 13.6) xx 1.602 xx 10^(-19)) = 1028 xx 10^(-10) m` `:. lambda = 1028 Å` `1` induced `lambda_(2) = (hc)/(E_(2) - E_(1)))` `E_(1) = -13.6 eV, E_(2) = -(13.6)/(4) eV` `:. lambda_(2)= (6.625 xx 10^(-34) xx 3 xx 10^(6))/((-(13.6)/(4) +13.6)xx1.602 xx 10^(-`19))` `:. induced lambda_(3) = (hc)/(E_(3) - E_(2))` `E_(1) = - 13.6 eV,E_(2) = - (13.6)/(4) eV` `E_(3) = -(13.6)/(9) eV` `:. lambda_(3) = (6.626 xx 10^(-34) xx 3 xx 10^(8))/((-(13.6)/(4) + (13.6)/(4)) xx1.602 xx 10^(-19))` `= 6568 xx 10^(-10) m = 6568 Å` |
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