Question

The maximum angular speed of the electron of a hydrogen atom in a stationary orbit is

• A. $6.2 \times 10^{15} \,rad \,s ^{-1}$
• B. $4.1 \times 10^{16} \,rad \,s ^{-1}$
• C. $24 \times 10^{10} \, rad \,s ^{-1}$
• D. $9.2 \times 10^{6} \,rad \,s ^{-1}$

Answer 1

Answer: (B)

Maximum angular speed will be in its ground state. Hence,
$\omega_{\max }=\frac{v_{1}}{r_{1}}=\frac{2.2 \times 10^{6}}{0.529 \times 10^{-10}}$
$=4.1 \times 10^{16} \,rad\, s ^{-1}$