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 quatum number n of the excited state is

Options

A. `sqrt((lambdaR)/(lambdaR-1))`
B. `sqrt((lambda)/(lambdaR-1))`
C. `sqrt((lambdaR^(2))/(lambdaR-1))`
D. `sqrt((lambdaR)/(lambda-1))`

Answer

A

Explanation

According to Rydberg.s formula, `(1)/(lambda)=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`
Here, `n_(f)=1, n_(i)=n`
`therefore (1)/(lambda)=R((1)/(1^(2))-(1)/(n^(2))) rArr (1)/(lambda)=R(1-(1)/(n^(2)))`
Multiplying equation (i) by `lambda` on both sides,
`1=lambdaR(1-(1)/(n^(2))) or (1)/(lambdaR)=1-(1)/(n^(2))`
or `(1)/(n^(2))=1-(1)/(lambdaR)" or "(1)/(n^(2))=(lambda-1)/(lambdaR) or `n=sqrt((lambdaR)/(lambdaR-1))`