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: Bohr model and magnetic field.
In Bohr’s model of the atom, the electron moves in discrete orbits around the nucleus, and each orbit corresponds to a particular energy level. The magnetic field at the centre of the orbit is related to the angular momentum of the electron. The magnetic field produced by the orbiting electron can be expressed as:
\[ B = \frac{\mu_0 m_e e v}{2r^2}, \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( m_e \) is the mass of the electron,
- \( e \) is the charge of the electron,
- \( v \) is the velocity of the electron,
- \( r \) is the radius of the orbit.

Step 2: Relating velocity and radius to quantum numbers.
From Bohr’s model, the radius of the orbit is quantized and given by:
\[ r = \frac{n^2 h^2 \varepsilon_0}{\pi m_e e^2}, \]
where \( n \) is the principal quantum number, \( h \) is Planck’s constant, and \( \varepsilon_0 \) is the permittivity of free space. The velocity of the electron is related to the radius by:
\[ v = \frac{2 \pi n h}{m_e r}. \]

Step 3: Magnetic field dependence on \( n \).
Substituting these expressions for \( v \) and \( r \) into the equation for \( B \), we find that the magnetic field at the centre is inversely proportional to the cube of the quantum number \( n \), i.e.,
\[ B \propto n^{-3}. \] Final Answer:
Thus, the magnetic field at the centre is proportional to \( n^{-3} \), and the correct answer is:
\[ \boxed{n^{-3}}. \]