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Calculate the wavelength and energy of radiation emitted for the electron transition from infinity to stationary state of the hydrogen atom |
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Answer» Correct Answer - A::B::C::D `(1)/(lambda) = R [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` ` = 109678 cm^(-1)[(1)/(1^(2)) - (1)/(oo^(2))]` `= 109678 cm^(-1)` `lambda = (1)/(109678 cm^(-1)) = 9.118 xx 10^(-6) cm` `E = (hc)/(lambda) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(9.118 xx 10^(-8)) = 2.178 xx 10^(18) J` |
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