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Magnetic moment due to the motion of the electron in the nth energy state of a hydrogen atom is proportional to

A

n

B

n⁰

C

n⁵

D

Step-by-Step Solution

According to the Bohr model, the orbital magnetic moment (μ\mu) of an electron is given by the product of the current (II) and the area of the orbit (AA).

  1. Current I=eT=ev2πrI = \frac{e}{T} = \frac{ev}{2\pi r}.
  2. Area A=πr2A = \pi r^2.
  3. Thus, μ=evr2\mu = \frac{evr}{2}. From Bohr's postulates, the angular momentum mvr=nh2πmvr = \frac{nh}{2\pi} . Substituting vr=nh2πmvr = \frac{nh}{2\pi m} into the magnetic moment equation: μ=e2(nh2πm)=n(eh4πm)=nμB\mu = \frac{e}{2} \left( \frac{nh}{2\pi m} \right) = n \left( \frac{eh}{4\pi m} \right) = n \mu_B. Therefore, μn\mu \propto n.
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