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Emission and Absorption Spectra

In the n th o...

Question

In the n^th orbit, the energy of an electron $E_{n} = - \frac{13.6}{n^{2}} e V$ for hydrogen atom. The energy required to take the electron from first orbit to second orbit will be

10.2 eV

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12.1 eV

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13.6 eV

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3.4 eV

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Solution

The correct option is A
10.2 eV

$E_{1 \to 2} = - 3.4 (13.6) = + 10.2 e V}$

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Similar questions

n^{t h}

E_{n} = - \frac{13.6}{n^{2}} eV

n^{t h}

E_{n} = - \frac{13.6}{n^{2}} e V

E_{n} = \frac{- 13.6}{n^{2}} e V

n t h

E_{n} = - \frac{13.6}{n^{2}} e V

13.6 e V

e V

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