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

Options

A. `sqrt((lambdaR)/(lambdaR-1))`
B. `sqrt((lambda)/(lambdaR-1))`
C. `sqrt((lambdaR^(2))/(lambda R-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`
`1/lambda=R (1l^(2)-1/n^(2)) rArr 1/lambda =R (1-1/n^(2)) ........(i)`
Multiplying equation (i) by `lambda` on both sides,
`1=lambda R(1-1/n^(2)) rArr 1/(lambda R)=1-1/n^(2)`
`rArr 1/n^(2)=1-(1)/(lambdaR) rArr 1/n^(2)=(lambdaR-1)/(lambdaR) rArr n= sqrt((lambdaR)/(lambdaR-1))`