If the energy of the second Bohr's orbit in H-atom is $- 328 k J$ $m o l^{- 1}$ , then what is the energy of the fourth Bohr orbit?

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

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

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

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

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Solution

The correct option is A $- 82 k J m o l^{- 1}$
If energy of an electron in the $n^{t h}$ orbit is $E_{n}$ , according to the Bohr's atomic model is
$E_{n} \propto \frac{Z^{2}}{n^{2}}$ ,
Thus, $\frac{E_{4}}{E_{2}} = \frac{2^{2}}{4^{2}} = \frac{1}{4}$

Now, $E_{2} = - 328 k J {m o l}^{- 1}$ ,
$E_{4} = \frac{1}{4} \times E_{2} = - 82 k J {m o l}^{- 1}$
Hence, option $A$ is correct.

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328 k J m o l^{- 1}

N a^{+}

- 389.4 {kJ mol}^{- 1}

C l^{-}

- 382.3 {kJ mol}^{- 1}

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N_{2} O

△_{f} H^{o} o f N_{2} O = 82 k J m o l^{- 1},

N \equiv N, 946 k J m o l^{- 1}; N = N, 418 k J m o l^{- 1};

O = O, 498 k J m o l^{- 1}; N = O, 607 k J m o l^{- 1}

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