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Mass Energy Equivalence

If the energy...

Question

If the energy of an electron in Hydrogen atom is given by expression, $- 1312 / n^{2} k J {m o l}^{- 1}$ , then the energy required to excite the electron from ground state to second orbit is

328 k J {m o l}^{- 1}

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656 k J {m o l}^{- 1}

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984 k J {m o l}^{- 1}

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1312 k J {m o l}^{- 1}

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Solution

The correct option is C $984 k J {m o l}^{- 1}$
$\begin{matrix} E = - 1312 (\frac{1}{4} - 1) = \frac{- 1312 \times 3}{4} = 328 \times 3 = 984 k J m o l^{- 1} H e n c e, o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

- \frac{1312}{n^{2}}

^{- 1}

2.46 L

1 atm

300 K

H - H

436 {kJ mol}^{- 1}

R = 0.082 {L atm K}^{- 1} {mol}^{- 1}

H -

= - 1312 {kJ mol}^{- 1}

- 328 {kJ mol}^{- 1}

n = 3

n = 2

E_{n} = \frac{- 1312}{n^{2}} k J m o l^{- 1}

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