1.

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

Answer»

`sqrt((lambdaR)/(lambdaR-1))`
`sqrt((LAMBDA)/(lambdaR-1))`
`sqrt((lambdaR^(2))/(lambda R-1))`
`sqrt((lambdaR)/(lambda-1))`

Solution :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))`


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