Question

The ratio of angular frequency (\(\omega\)) of an electron to the total energy ( E) of an electron in the first orbit of hydrogen atom is ( \(h\) is the Planck's constant)

• A. \(-h/4\pi\)
• B. \(-4\pi /h\)
• C. \(-2\pi /h\)
• D. \(-h/2\pi\)

Hint:

Use the Bohr quantisation of angular momentum and the fact that total energy equals minus the kinetic energy.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In a Bohr orbit the total energy is \(E=-K\), where \(K=\dfrac{1}{2}mv^{2}\) is the kinetic energy.

Step 2: Express with angular momentum
\(K=\dfrac{1}{2}mr^{2}\omega^{2}=\dfrac{1}{2}L\omega\), so \(E=-\dfrac{1}{2}L\omega\).

Step 3: Calculation
For the first orbit \(L=\dfrac{h}{2\pi}\).
\(E=-\dfrac{h\omega}{4\pi}\), so \(\dfrac{\omega}{E}=-\dfrac{4\pi}{h}\).

Final Answer:
The ratio of \(\omega\) to \(E\) in the first orbit is \(-4\pi/h\). \[ \boxed{\text{(B) }-\dfrac{4\pi}{h}} \]