Q55 / 200 · NEET 2005 · Chemistry
The energy of second Bohr orbit of the hydrogen atom is $-328 \mathrm{~kJ} \mathrm{~mol}^{-1}$; hence the energy of fourth Bohr orbit would be
- A$-41 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- Correct answerB$-82 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- C$-164 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$-1312 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$\mathrm{Z}=1$ for hydrogen ; $n=2$
$$
\begin{aligned}
& \mathrm{E}_n=-k\left(\frac{\mathrm{Z}}{n}\right)^2 \\
& \mathrm{E}_2=\frac{-k \times 1}{4} \\
& \mathrm{E}_2=-328 \mathrm{KJ} \mathrm{mol}^{-1} ; k=4 \times 328 \\
& \mathrm{E}_4=\frac{-k \times 1}{16} \\
& \mathrm{E}_4=-4 \times 328 \times \frac{1}{16} \\
& \mathrm{E}_4=-82 \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$$
Related Theory
The total mechanical energy of an electron in a Bohr orbit is the sum of its kinetic and potential energies. The lowest energy is called the ground state.
Caution
It is given that the energy of the electron in the first Bohr $\mathrm{H}$-atom is $-13.6 \mathrm{eV}$
Hence, the correct answer is option: 'B'.