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Calculate the energy of radiation emitte...

Calculate the energy of radiation emitted for the electronic transition from infinity to ground state for hydrogen atom. Given `c = 3xx10^(8)" ms"^(-1), R_(H) = 1.09678 xx 10^(7) m^(-1), h = 6.6256 xx 10^(-34)" J s"^(-1)`

A

`2.18 xx 10^(-18)J`

B

`3.25 xx 10^(-18)J`

C

`4.05 xx 10^(-18)J`

D

`2.39 xx 10^(-18)J`

Text Solution

Verified by Experts

The correct Answer is:
A
According to Rydberg equation, we have
`barv=(1)/(lambda) =RZ^(2)[(1)/(n_(1)^(2)) -(1)/(n_(2)^(2))]` where `Z=1, R=1.09678 xx 10^(7)m^(-1), n_(1)=1` and `n_(2)=oo`
Therefore, `(1)/(lambda)= 1.09678xx10^(7) [(1)/(1)-(1)/(oo)] or lambda =(1)/(1.09678xx10^(7)) =0.911 xx 10^(-7) = 9.11 xx 10^(-8)m`
The energy is given by `E=(hc)/(lambda)=(6.6256xx10^(-34) xx 3xx10^(8))/(9.11 xx 10^(-8))=2.18 xx 10^(-18)J`
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