If there are two de-Broglie waves formed in an orbit of H-atom then the energy of electron in that orbit is.

- 3.4

e V

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- 13.6

e V

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- 9.06

e V

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1.51

e V

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Solution

The correct option is C $- 3.4$ $e V$
$\begin{matrix} 2 π r = 2 \cdot \frac{h}{m v} \Rightarrow m v r = \frac{h}{π} = 2 (\frac{h}{2 π}) \Rightarrow n = 2 ∴ E = - 13.6 \times \frac{1}{4} = - 3.4 e v H e n c e, o p t i o n A i s c o r r e c t a n s w e r . \end{matrix}$

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If there are two de-Broglie waves formed in an orbit of H-atom then the energy of electron in that orbit is.

The correct option is C −3.4 eV2πr=2⋅hmv⇒mvr=hπ=2(h2π)⇒n=2∴E=−13.6×14=−3.4evHence,optionAiscorrectanswer.

The correct option is C $- 3.4$ $e V$
$\begin{matrix} 2 π r = 2 \cdot \frac{h}{m v} \Rightarrow m v r = \frac{h}{π} = 2 (\frac{h}{2 π}) \Rightarrow n = 2 ∴ E = - 13.6 \times \frac{1}{4} = - 3.4 e v H e n c e, o p t i o n A i s c o r r e c t a n s w e r . \end{matrix}$