Question

The angular velocity of an electron occupying the second Bohr orbit of $H{e}^{+}$ ion is: (in $s^{-1}$)

• A. $2.06\times {10}^{16}$
• B. $3.06\times {10}^{16}$
• C. $1.06\times {10}^{18}$
• D. $2.06\times {10}^{17}$

Answer 1

The correct option is A $2.06\times {10}^{16}$

Velocity of an electron in $H{e}^{+}$ ion in an orbit (v)
$=2.18\frac{Z}{n}\times {10}^{6}m/s$
Radius of $H{e}^{+}$ ion in an orbit $\left(r_n\right)=5.29\frac{n^2}{Z}\times {10}^{-11}m$
Angular velocity (w) = $\frac{v}{r_n}=\frac{2.18}{5.29}\times \frac{Z^2}{n^3}\times {10}^{17}\approx 2.06\times {10}^{16}{\text{sec}}^{-1}$