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Calculate the wavelength and energy of radiation emitted for the electron transition from infinite `(oo)` to first 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» `(1)/(lamda)=R[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` `n_(1)=1 and n_(2)=infty` `(1)/(lamda)=R[(1)/(1^(2))-(1)/((infty)^(2))]=R` or `lamda=(1)/(R)=(1)/(1.09678xx10^(7))=9.11xx10^(-8)m` We know that, `E=hv=h.(c)/(lamda)=6.6256xx10^(-34)xx(2.9979xx10^(8))/(9.11xx10^(-8))` `=2.17xx10^(-18)J`. |
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