The ratio of magnitude of energies of electron in first and second excited states of hydrogen atom is:

1 : 4

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4 : 9

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9 : 4

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4 : 1

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Solution

The correct option is C $9 : 4$
The energy of the $n^{t h}$ state of the single electron system is given by,

$E_{n} = (\frac{Z^{2}}{n^{2}}) (- 13.6 eV)$

For Hydrogen atom $Z = 1$ .

As for first excited state, $n_{2} = 2$ and for second excited state, $n_{3} = 3$

$\Rightarrow \frac{E_{2}}{E_{3}} = {(\frac{n_{3}}{n_{2}})}^{2} = {(\frac{3}{2})}^{2} = \frac{9}{4}$

Hence, option $(C)$ is correct.

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The ratio of magnitude of energies of electron in first and second excited states of hydrogen atom is:

The correct option is C 9:4The energy of the nth state of the single electron system is given by, En=(Z2n2)(−13.6 eV) For Hydrogen atom Z=1. As for first excited state, n2=2 and for second excited state, n3=3 ⇒E2E3=(n3n2)2=(32)2=94 Hence, option (C) is correct.