Question:
In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in \(n^{\mathrm{th}}\) excited state, is :
In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in \(n^{\mathrm{th}}\) excited state, is :
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Magnetic moment of a current loop is \(\mu = iA\); relate it to the orbital angular momentum \(L = mvr = nh/2\pi\).
Updated On: Oct 31, 2026
- \(\left(\dfrac{e}{2m}\right)\dfrac{n^2h}{2\pi}\)
- \(\left(\dfrac{e}{m}\right)\dfrac{nh}{2\pi}\)
- \(\left(\dfrac{e}{2m}\right)\dfrac{nh}{2\pi}\)
- \(\left(\dfrac{e}{m}\right)\dfrac{n^2h}{2\pi}\)
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The Correct Option is C
Solution and Explanation
Orbiting electron forms a current \(i = \dfrac{e}{T} = \dfrac{e v}{2\pi r}\).
Magnetic moment \(\mu = iA = \dfrac{e v}{2\pi r}\cdot\pi r^2 = \dfrac{e v r}{2}\).
Multiply and divide by m: \(\mu = \dfrac{e}{2m}(m v r) = \dfrac{e}{2m}L\).
Bohr's quantisation: \(L = \dfrac{nh}{2\pi}\).
\(\mu = \left(\dfrac{e}{2m}\right)\dfrac{nh}{2\pi}\).
Answer: (3) \(\left(\dfrac{e}{2m}\right)\dfrac{nh}{2\pi}\)
Magnetic moment \(\mu = iA = \dfrac{e v}{2\pi r}\cdot\pi r^2 = \dfrac{e v r}{2}\).
Multiply and divide by m: \(\mu = \dfrac{e}{2m}(m v r) = \dfrac{e}{2m}L\).
Bohr's quantisation: \(L = \dfrac{nh}{2\pi}\).
\(\mu = \left(\dfrac{e}{2m}\right)\dfrac{nh}{2\pi}\).
Answer: (3) \(\left(\dfrac{e}{2m}\right)\dfrac{nh}{2\pi}\)
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