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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 |
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Answer» `sqrt((lambdaR)/(lambdaR-1))` `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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