Question

Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the 2nd and 4th orbits of a hydrogen atom.

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
An electron orbiting a nucleus behaves like a circular current loop.
This loop generates a magnetic field at the center of the orbit.
By expressing the magnetic field in terms of the principal quantum number \(n\), we can easily find the ratio for different orbits.
Step 2: Key Formula or Approach:
The magnetic field at the center of a circular current loop is \(B = \frac{\mu_0 I}{2r}\).
The equivalent current is \(I = \frac{e}{T} = \frac{ev}{2\pi r}\).
Substituting \(I\) gives \(B = \frac{\mu_0 ev}{4\pi r^2}\).
From Bohr's model, velocity \(v \propto \frac{1}{n}\) and radius \(r \propto n^2\).
Step 3: Detailed Explanation:
Substitute the proportionalities of \(v\) and \(r\) into the magnetic field equation:
\[ B \propto \frac{v}{r^2} \] \[ B \propto \frac{(1/n)}{(n^2)^2} \] \[ B \propto \frac{1/n}{n^4} = \frac{1}{n^5} \] So, the magnetic field is inversely proportional to the fifth power of the principal quantum number.
We need the ratio of the magnetic fields in the 2nd (\(n_1 = 2\)) and 4th (\(n_2 = 4\)) orbits.
\[ \frac{B_2}{B_4} = \frac{(n_4)^5}{(n_2)^5} \] \[ \frac{B_2}{B_4} = \left(\frac{4}{2}\right)^5 \] \[ \frac{B_2}{B_4} = (2)^5 = 32 \] Step 4: Final Answer:
The ratio of the magnetic fields is \(32\).