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Calculate the wavelength and energy for radiation emitted for the electron transition from infinite `(oo)` to stationary state of the hydrogen atom `R = 1.0967 xx 10^(7) m^(-1), h = 6.6256 xx 10^(-34) J s ` and `c = 2.979 xx 10^(8) m s^(-1)` |
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Answer» Correct Answer - A::B::C::D `n_(1) = 1,n_(2) = oo,(1)/(lambda) = 1.09678 xx 10^(7) [(1)/(1^(2)) - (1)/(oo^(2))]` `lambda = 9.11 xx 10^(6) m ` `lambda = 9.11 xx 10^(-6) m` `E = hv = b(c )/(lambda) = Rhc (since (1)/(lambda) = R)` ` = 1.09678 xx 10^(7) xx 6.62 xx 10^(-34) xx 3 xx 10^(8) m` `= 217.9 xx 10^(-20) J` `217.9 xx 10^(-23) kJ` |
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