Question

What is the frequency \( \nu \) of the electron in Bohr's first orbit of radius \( r \) of the hydrogen atom?

• A. \( \nu = \dfrac{e^2}{4\pi \varepsilon_0 h r} \)
• B. \( \nu = \dfrac{e^2}{2\pi \varepsilon_0 h r^2} \)
• C. \( \nu = \dfrac{e^2}{4\pi \varepsilon_0 h r^2} \)
• D. \( \nu = \dfrac{e^2}{2\pi \varepsilon_0 h r} \)

Hint:

In Bohr model problems, combine Coulomb force with quantization condition \( mvr = \frac{nh}{2\pi} \), and use \( \nu = \frac{v}{2\pi r} \) for frequency.

Answer 1

The Correct Option is A

Step 1: Use Coulomb force as centripetal force.
In Bohr’s model, the electrostatic force provides the centripetal force:
\[ \frac{1}{4\pi \varepsilon_0} \cdot \frac{e^2}{r^2} = \frac{mv^2}{r} \]

Step 2: Simplify the equation.
\[ \frac{1}{4\pi \varepsilon_0} \cdot \frac{e^2}{r} = mv^2 \]

Step 3: Use Bohr’s quantization condition.
Angular momentum is quantized:
\[ mvr = \frac{nh}{2\pi} \] For first orbit, \( n = 1 \): \[ mvr = \frac{h}{2\pi} \]

Step 4: Express velocity \( v \).
\[ v = \frac{h}{2\pi mr} \]

Step 5: Substitute into centripetal equation.
Substitute \( v \) in
Step 2:
\[ \frac{1}{4\pi \varepsilon_0} \cdot \frac{e^2}{r} = m \left( \frac{h}{2\pi mr} \right)^2 \]

Step 6: Simplify to get velocity relation.
After simplification:
\[ v = \frac{e^2}{2\varepsilon_0 h} \]

Step 7: Find frequency \( \nu \).
Frequency is:
\[ \nu = \frac{v}{2\pi r} \] \[ \nu = \frac{e^2}{4\pi \varepsilon_0 h r} \]
Hence, option (A) is correct.