The ratio of energies of first two excited states hydrogen atom is ?

3/1

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1/4

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4/9

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9/4

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Solution

The correct option is D 9/4
Ground state or $n = 1$ state has enegy $E_{1} = - 13.6 e V$
First excited state means $n = 2$ so $E_{2} = E_{1} / 2^{2} = - 3.4 e V$
Second excited state means $n = 3$ so $E_{3} = - 13.6 / 3^{2} = - 1.51 e V$
The ratio is $\frac{- 3.4}{- 1.51} = 2.25 = 9 / 4$ or simply $\frac{3^{2}}{2^{2}}$
Option D is correct.

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The ratio of energies of first two excited states hydrogen atom is ?

The correct option is D 9/4Ground state or n=1 state has enegy E1=−13.6eVFirst excited state means n=2 so E2=E1/22=−3.4eVSecond excited state means n=3 so E3=−13.6/32=−1.51eVThe ratio is −3.4−1.51=2.25=9/4 or simply 3222Option D is correct.