Question

The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is (\(h\) = Planck's constant, \(m\) is the mass of electron and \(r\) is the radius of the orbit.)

• A. \(h/4π^2mr^2\)
• B. \(h/4π^2mr\)
• C. \(h/4πmr\)
• D. \(h/4π^2m^2r^2\)

Hint:

Frequency is $\frac{v}{2\pi r}$, and Bohr gives $mvr=\frac{h}{2\pi}$.

Answer 1

The Correct Option is A

Step 1: Use angular momentum and angular speed
$L=I\omega=mr^2\omega=\frac{h}{2\pi}$, so $\omega=\frac{h}{2\pi mr^2}$.
$f=\frac{\omega}{2\pi}=\frac{h}{4\pi^2mr^2}$.

Final Answer:
Option (A). \[ \boxed{\text{(A)}} \]