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Hydrogen atom from excited state comes t...

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength `lambda`. If R is the Rydberg constant, then the principal quantum number n of the excited state is

A

`sqrt((lamda R)/(lamda R-1))`

B

`sqrt((lamda)/(lamda R-1))`

C

`sqrt((lamda R^(2))/(lamda R-1))`

D

`sqrt((lamda R)/(lamda-1))`

Text Solution

Verified by Experts

The correct Answer is:
A
`(1)/(lamda) = R ((1)/(n_(1)^(2))- (1)/(n_(2)^(2)))`
`rArr (1)/(lamda) = R (1- (1)/(n^(2))) rArr 1- (1)/(n^(2))= (1)/(lamda R)`
`rArr (1)/(n^(2))= 1- (1)/(lamda R) rArr n= sqrt((lamda R)/(lamda R-1))`
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