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Q. The ratio of energies of hydrogen atom in its first excited state to third excited state is

Atoms

Solution:

Energy in $n^{\text {th }}$ excited state $=\frac{-13.6}{(n+1)^{2}}$
Energy in $1^{\text {st }}$ excited state $=\frac{-13.6}{4}$
Energy in $3^{\text {ds }}$ excited state $=\frac{-13.6}{16}$
Ratio $E_{1}: E_{3}=4: 1$