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JEE Main 20139 AprilOnline
Exam
JEE Main
Year
2013
Date
9 April
Mode
Online
Subject
Physics
Chapter
Atomic Physics

JEE Main 2013 · 9 April · Online · Physics · Atomic Physics

In the Bohr model an electron moves in a circular orbit around the...

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 nth n^{\text{th }} excited state, is :

Options

  1. A(e2mn2h2π)\left(\frac{e}{2 m} \frac{n^2 h}{2 \pi}\right)
  2. B(em)nh2π\left(\frac{e}{m}\right) \frac{n h}{2 \pi}
  3. C(e2m)nh2π\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}
  4. D(em)n2h2π\left(\frac{e}{m}\right) \frac{n^2 h}{2 \pi}

Correct answer

C. (e2m)nh2π\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}

Step-by-step solution

Magnetic moment of the hydrogen atom, when the electron is in nth \mathrm{n}^{\text{th }} excited state, i.e., n′=(n+1)n^{\prime}=(n+1) As magnetic moment Mn=\mathrm{M}_{\mathrm{n}}=\mathrm{I}n_{\mathrm{n}}\mathrm{A}=\mathrm{i}n(πrn2)_{\mathrm{n}}\left(\pi \mathrm{r}_{\mathrm{n}}^2\right)in=eVn=mz2e54ε02n3h3rn=n2h24π2kzme2(k=14πϵ0)\begin{aligned} & i_n=e V_n=\frac{m z^2 e^5}{4 \varepsilon_0^2 n^3 h^3} \\ & r_n=\frac{n^2 h^2}{4 \pi^2 k z m e^2}\left(k=\frac{1}{4 \pi \epsilon_0}\right) \end{aligned}$SolvingwegetmagneticmomentofthehydrogenatomforSolving we get magnetic moment of the hydrogen atom for\mathrm{n}^{\text{th }}excitedstateexcited stateM_{n^{\prime}}=\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$
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