An electron enters a uniform magnetic field of flux density $1.2\ \text{Wb/m}^2$. The energy difference (in eV) between electrons having spins parallel and anti-parallel to the field is:
(Given: $\mu_B = 9.3 \times 10^{-24}\ \text{J/T}$)

Question

An electron is revolving around a proton in an orbit of radius r with a speed v. Obtain an expression for the magnetic moment associated with the electron.

Answer 1

Step 1: Think of this as Zeeman-type splitting. Each spin state shifts by an energy $\mu_B B$; the parallel state is lowered and the anti-parallel state raised (or vice versa), giving a gap of $2\mu_B B$.

Step 2: Keep everything in eV from the start. First express the Bohr magneton in eV/T:
\[\mu_B = \frac{9.3 \times 10^{-24}\ \text{J/T}}{1.6 \times 10^{-19}\ \text{J/eV}} = 5.8125 \times 10^{-5}\ \text{eV/T}\]
Step 3: The gap is
\[\Delta U = 2\mu_B B = 2 \times 5.8125 \times 10^{-5} \times 1.2\ \text{eV}\]
Step 4: Evaluate:
\[\Delta U = 2 \times 6.975 \times 10^{-5} = 13.95 \times 10^{-5}\ \text{eV}\]
The same numerical answer confirms option (B).
\[\boxed{\Delta U = 13.95 \times 10^{-5}\ \text{eV}}\]

An electron enters a uniform magnetic field of flux density \(1.2\ \text{Wb/m}^2\). The energy difference (in eV) between electrons having spins parallel and anti-parallel to the field is:
(Given: \(\mu_B = 9.3 \times 10^{-24}\ \text{J/T}\))

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The Correct Option is B

Solution and Explanation

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An electron enters a uniform magnetic field of flux density \(1.2\ \text{Wb/m}^2\). The energy difference (in eV) between electrons having spins parallel and anti-parallel to the field is:(Given: \(\mu_B = 9.3 \times 10^{-24}\ \text{J/T}\))