The energy of second Bohr orbit of hydrogen atom is -328kJ.mol-1 , hence energy of fourth Bohr orbit would be

# The energy of second Bohr orbit of hydrogen atom is $-$328$\mathrm{kJ}.{\mathrm{mol}}^{-1}$ , hence energy of fourth Bohr orbit would be

1. A

$-41\mathrm{kJ}.{\mathrm{mol}}^{-1}$

2. B

$-82\mathrm{kJ}.{\mathrm{mol}}^{-1}$

3. C

$-164\mathrm{kJ}.{\mathrm{mol}}^{-1}$

4. D

$-1312\mathrm{kJ}.{\mathrm{mol}}^{-1}$

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### Solution:

The energy of hydrogen atom in Bohr’s orbit is

${\mathrm{E}}_{\mathrm{n}}=\frac{-1312}{{\mathrm{n}}^{2}}\mathrm{kJ}.{\mathrm{mol}}^{-1}$

For hydrogen atom, the ratio of energy between two different orbits is

Energy of Bohr’s second orbit ( n=2 ) is ${\mathrm{E}}_{2}=-328\mathrm{kJ}.{\mathrm{mol}}^{-1}.$

Energy of Bohr’s fourth orbit ( n = 4 ) is ${\mathrm{E}}_{4}=?$

From the equation,

${\mathrm{E}}_{4}=\frac{{2}^{2}}{{4}^{2}}×{\mathrm{E}}_{2}=\frac{{2}^{2}}{{4}^{2}}×\left(-328\mathrm{kJ}.{\mathrm{mol}}^{-1}\right)$

$\therefore {\mathrm{E}}_{4}=-82\mathrm{kJ}.{\mathrm{mol}}^{-1}$

As we going away from nucleus magnitude of energy increases.

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