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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)`

Text Solution

Verified by Experts

`(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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