Question

The angular momentum of an electron in a Bohr's orbit of $H-$ atom is $4.2178\times 10^{- 34}kg-m^{2}/s$ . Number of spectral line emitted when electrons falls from this level to next lower level :-

Answer 1

Angular momentum
$mvr=\frac{nh}{2 \pi }=4.2178\times 10^{- 34}$
$\therefore n=\frac{4 . 2178 \times 10^{- 34} \times 2 \times 3 . 14}{6 . 62 \times 10^{- 34}}=4$
when $e^{\ominus}$ falls from $n=4$ to next lower level
$n=3.$ Then only one spectral line will be obtained.