Question

Consider Bohr's model of a H-atom. If magnetic field at center due to electron in 2\(^n\) orbit is \( B_1 \), and magnetic field due to electron in 4\(^t\) orbit is \( B_2 \). Find \( \frac{B_1}{B_2} \).

Hint:

In Bohr's model, the magnetic field is inversely proportional to the square of the radius. The radius increases as the square of the orbit number.

Answer 1

Correct Answer: 64

Step 1: Expression for magnetic field in Bohr's model.
The magnetic field at the center due to an electron in orbit \( n \) is given by the formula: \[ B = \frac{{\mu_0 \cdot e \cdot v}}{{2 \cdot r^2}} \] where \( \mu_0 \) is the permeability of free space, \( e \) is the charge of the electron, \( v \) is the velocity of the electron, and \( r \) is the radius of the orbit.
Step 2: Relationship between radius and orbit number.
The radius of the orbit in Bohr's model is given by: \[ r_n = \frac{{n^2 \cdot h^2}}{{4 \pi^2 \cdot m_e \cdot e^2}} \quad \text{where} \quad n \text{ is the orbit number}. \]
Step 3: Magnetic field due to electron in orbit \( n \).
For the 2nd and 4th orbits, the magnetic field will be: \[ B_1 \propto \frac{1}{r_2^2}, \quad B_2 \propto \frac{1}{r_4^2}. \]
Step 4: Relationship between \( B_1 \) and \( B_2 \).
From the formula for the radius: \[ r_2 = \frac{2^2 \cdot h^2}{4 \pi^2 \cdot m_e \cdot e^2}, \quad r_4 = \frac{4^2 \cdot h^2}{4 \pi^2 \cdot m_e \cdot e^2}. \] Thus, we have: \[ B_1 \propto \frac{1}{r_2^2}, \quad B_2 \propto \frac{1}{r_4^2}. \] Using the relation between the radii, we can find: \[ \frac{B_1}{B_2} = \frac{r_4^2}{r_2^2} = \frac{(4^2)^2}{(2^2)^2} = \frac{16}{4} = 64. \]
Step 5: Final answer.
Thus, the ratio \( \frac{B_1}{B_2} \) is: \[ \boxed{64}. \]