Assuming the atom is in the ground state, the expression for the magnetic field at a point nucleus in hydrogen atom due to circular motion of electron is ( μ₀ = permeability of free space, ε₀ = permittivity of free space, m = mass of electron, e = electronic charge) (a) (μ₀ e² m²)/(8ε₀ h²) (b) (μ₀ e⁷ π m²)/(8ε₀³ h⁵) (c) (μ₀ e⁵ π m)/(6ε₀² h²) (d) (μ₀ e² π m²)/(8ε₀ h²)

Answer: (b). B = (μ₀ I)/(2r) with I = (ev)/(2π r), so B = (μ₀ e v)/(4π r²) Ground state: v = (e²)/(2ε₀ h), r = (ε₀ h²)/(π m e²) B = (μ₀ e)/(4π)·(e²)/(2ε₀ h)·(π² m² e⁴)/(ε₀² h⁴) B = (μ₀ e⁷ π m²)/(8ε₀³ h⁵), so the answer is (b).

Assuming the atom is in the ground state, the expression for the… | MHT CET 2026 (12 May, Shift 2)

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Bohr Model: Orbits, Speed & Angular Momentum

Assuming the atom is in the ground state, the expression for the magnetic field at a point nucleus in hydrogen atom due to circular motion of electron is ( $\mu_0$ = permeability of free space, $\epsilon_0$ = permittivity of free space, $m$ = mass of electron, $e$ = electronic charge)

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Answer: (b)

$B = \frac{\mu_0 I}{2r}$ with $I = \frac{ev}{2\pi r}$ , so $B = \frac{\mu_0 e v}{4\pi r^2}$
Ground state: $v = \frac{e^2}{2\varepsilon_0 h}$ , $r = \frac{\varepsilon_0 h^2}{\pi m e^2}$
$B = \frac{\mu_0 e}{4\pi}\cdot\frac{e^2}{2\varepsilon_0 h}\cdot\frac{\pi^2 m^2 e^4}{\varepsilon_0^2 h^4}$
$B = \frac{\mu_0 e^7 \pi m^2}{8\varepsilon_0^3 h^5}$ , so the answer is (b).

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