The energy of an electron in an excited hydrogen atom is given as -3.…

The correct answer is Option (3) → \( 2.1\times10^{-34}\,\mathrm{J\,s} \)

Bohr’s quantisation condition says the angular momentum in the \( n \)th orbit is \( L = nh/2\pi \). First find \( n \) from the energy.

    \( \displaystyle E_n = -\frac{13.6}{n^{2}}\,\mathrm{eV} \)

    \( \displaystyle -3.4 = -\frac{13.6}{n^{2}} \Rightarrow n^{2} = 4 \Rightarrow n = 2 \)

    \( \displaystyle L = \frac{nh}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi} \)

    \( \displaystyle L = \frac{6.63\times10^{-34}\,\mathrm{J\,s}}{3.14} \approx 2.1\times10^{-34}\,\mathrm{J\,s} \)