Q.An electron in an atom is revolving round the nucleus in a circular orbit of radius $5.3 \times 10^{-11}$ m with a speed of $3 \times 10^6$ m/s. Find the angular momentum of electron.

Bohr's model was designed for and works only for single-electron (hydrogen-like) systems; it breaks down once more than one electron interacts within the atom.

One key limitation of Bohr's atomic model: it successfully explains the line spectrum of hydrogen (and other hydrogen-like single-electron ions such as He$^+$, Li$^{2+}$), but it fails to explain the spectra of atoms with more than one electron, because it does not account for the electron-electron interactions/repulsions present in multi-electron atoms. (Other valid limitations include: it does not explain the relative intensities of spectral lines, the fine structure (splitting) of spectral lines, or the Zeeman/Stark effects, and it arbitr …

Bohr's third (quantisation) postulate as a formula.

Bohr's postulate of quantisation of angular momentum states that the electron can revolve only in those orbits for which its orbital angular momentum is an integral multiple of $h/2\pi$: …

Bohr's quantization condition gives $L=nh/2\pi$; for $n=3$, $L=3h/2\pi$.

According to Bohr's second postulate, the angular momentum of an electron in the $n$-th orbit is quantized: …

Bohr's angular momentum quantization gives L = nh/2π; the change from n=3 to n=1 is ΔL = 2h/2π = h/π.

By Bohr's postulate, $L_n = \dfrac{nh}{2\pi}$. For the transition from $n=3$ to $n=1$:

$\Delta L = L_3 - L_1 = \frac{3h}{2\pi} - \frac{h}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi}$ …