Question

Calculate the angular velocity of an electron occupying the second Bohr orbit of $H{e}^{+}$ion.

• A. $2.067\times {10}^{16}$
• B. $2.4\times {10}^{-4}$
• C. $1.65\times {10}^{17}$
• D. $4.9\times {10}^{-4}$

Answer 1

Answer: (A)

$2.067\times {10}^{16}$

Velocity of an electron in $H{e}^{+}$ ion in an orbit$=\frac{2\pi Z{e}^2}{nh}=u$
Radius of $H{e}^{+}$ ion in an orbit$=\frac{n^2{h}^2}{4{\pi }^{2}m{e}^{2}Z}=r$
Angular velocity $\left(\omega \right)=\frac{u}{r}=\frac{8{\pi }^3{Z}^{2}m{e}^4}{n^3{h}^3}$
$=\frac{8\times (22/7{)}^3\times (2{)}^2\times \times (9.108\times {10}^{-28})\times (4.803\times {10}^{-10}{)}^4}{(2{)}^3\times (6.626\times {10}^{-27}{)}^3}$
$=2.067\times {10}^{16}se{c}^{-1}$