Question

An electron moves in Bohr orbit. The magnetic field at the centre is proportional to

• A. \( n^{-2} \)
• B. \( n^{-3} \)
• C. \( n^{-4} \)
• D. \( n^{-5} \)

Hint:

In Bohr’s model, the magnetic field at the centre of the orbit is related to the angular momentum, and it depends on the quantum number \( n \) raised to the power of -3.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
An electron in a Bohr orbit acts like a current loop. We need to find how the magnetic field at the centre of this circular path scales with the principal quantum number \( n \).
Step 2: Key Formula or Approach:
1. Magnetic field at center of a loop: \( B = \frac{\mu_0 I}{2r} \).
2. Current \( I = \frac{e}{T} = \frac{ev}{2\pi r} \).
3. In Bohr's model: Velocity \( v \propto \frac{1}{n} \) and Radius \( r \propto n^2 \).
Step 3: Detailed Explanation:
Substitute the current into the magnetic field formula:
\[ B \propto \frac{I}{r} \propto \frac{v/r}{r} = \frac{v}{r^2} \]
Now substitute the proportionalities for \( v \) and \( r \):
\[ B \propto \frac{1/n}{(n^2)^2} = \frac{1/n}{n^4} = \frac{1}{n^5} \]
\[ B \propto n^{-5} \]
Step 4: Final Answer:
The magnetic field at the centre is proportional to \( n^{-5} \).