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

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

$\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{n}}_2^2}{{\mathrm{n}}_1^2}; \mathrm{Here} {\mathrm{E}}_1 \mathrm{and} {\mathrm{E}}_2 \mathrm{are} \mathrm{the} \mathrm{energies} \mathrm{of} \mathrm{the} \mathrm{orbits} \mathrm{of} {\mathrm{n}}_{1 }\mathrm{and} {\mathrm{n}}_2.$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,

$\frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{n}}_2^2}{{\mathrm{n}}_1^2}; \mathrm{then} \frac{{\mathrm{E}}_4}{{\mathrm{E}}_2}=\frac{2^2}{4^2}$

${\mathrm{E}}_4=\frac{2^2}{4^2}\times {\mathrm{E}}_2=\frac{2^2}{4^2}\times \left(-328\mathrm{kJ}.{\mathrm{mol}}^{-1}\right)$

$\therefore {\mathrm{E}}_4=-82\mathrm{kJ}.{\mathrm{mol}}^{-1}$

As we going away from nucleus magnitude of energy increases.