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Eelectron in hydrogen atom first jumps f...

Eelectron in hydrogen atom first jumps from third excited state to second excited state and tehn from second excited to the first excited state. The rartio of the wavelength `lambda_(1):lambda_(2)` emitted in the two cases is

A

`27//5`

B

`20//7`

C

`7//5`

D

`27//20`

Text Solution

Verified by Experts

The correct Answer is:
B
Wavelength observed from transition of `n_(i) " to " n_(f) " is " lamda=(1)/([(1)/(n_(f)^(2))-(1)/(n_(i)^(2))])`
For `lamda_(1), n_(i)=4, n_(f)=3`
For `lamda_(2),n_(i)=3,n_(f)=2`
We get
`lamda_(1) : lamda_(2)=20 : 7`
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