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Q. The energy of an electron in the excited state of $H$-atom is $- 1.5\, eV$, then according to Bohrs model, its momentum will be:

Rajasthan PMTRajasthan PMT 2000

Solution:

$\left(n^{2}=\frac{13.6}{1.5} \Rightarrow n^{2}=9\right)$
$\therefore n=3$
For $1.5\, eV , n=3$
Angular momentum
$=n \frac{h}{2 \pi} $
$=\frac{3 \times 6.6 \times 10^{-34}}{2 \times 3.14}$
$=3.15 \times 10^{-34} J - sec$