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: Frequency
Revolutions per second \(f=\frac{v}{2\pi r}\).

Step 2: Speed from quantisation
For the first orbit \(mvr=\frac{h}{2\pi}\), so \(v=\frac{h}{2\pi mr}\).

Step 3: Substitute
\(f=\frac{h}{2\pi mr}\times\frac{1}{2\pi r}=\frac{h}{4\pi^2mr^2}\). Option (A).

Final Answer:
The frequency is \(\frac{h}{4\pi^2mr^2}\), option (A). \[ \boxed{\text{(A)}} \]