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Atoms and Nuclei question

2017 · 8 Apr · Shift 1 · Q75
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Atoms and Nuclei question

2017 · 8 Apr · Shift 1 · Q75

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
According to Bohr’s theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the nth orbit is proportional to : (n = principal quantum number)
  1. A
    n−4{n^{ - 4}}n−4
  2. B
    n−5{n^{ - 5}}n−5
  3. C
    n −-− 3
  4. D
    n −-− 2
View written solutionFree

Correct answer: B

  1. Magnetic field at the centre due to an orbiting electron

An electron revolving in a circular Bohr orbit constitutes a current loop.

For a circular loop of radius rrr, the magnetic field at the centre is

B=μ0I2rB=\frac{\mu_0 I}{2r}B=2rμ0​I​

where III is the equivalent current.

  1. Current due to revolving electron

If the electron completes one revolution in time period TTT, then

I=eTI=\frac{e}{T}I=Te​

Also,

T=2πrvT=\frac{2\pi r}{v}T=v2πr​

So,

I=eT=ev2πrI=\frac{e}{T}=\frac{ev}{2\pi r}I=Te​=2πrev​

Hence,

B=μ02r⋅ev2πr=μ0ev4πr2B=\frac{\mu_0}{2r}\cdot \frac{ev}{2\pi r} =\frac{\mu_0 ev}{4\pi r^2}B=2rμ0​​⋅2πrev​=4πr2μ0​ev​

Thus,

B∝vr2B \propto \frac{v}{r^2}B∝r2v​
  1. Use Bohr model dependence on nnn

For hydrogen atom in the nnnth orbit,

rn∝n2r_n \propto n^2rn​∝n2

and

vn∝1nv_n \propto \frac{1}{n}vn​∝n1​

Substituting into

B∝vr2B \propto \frac{v}{r^2}B∝r2v​

we get

Bn∝1/n(n2)2=1n5B_n \propto \frac{1/n}{(n^2)^2} =\frac{1}{n^5}Bn​∝(n2)21/n​=n51​

Therefore,

Bn∝n−5B_n \propto n^{-5}Bn​∝n−5
  1. Option check
  • A: n−4n^{-4}n−4 ❌
  • B: n−5n^{-5}n−5 ✅
  • C: n−3n-3n−3 ❌
  • D: n−2n-2n−2 ❌

So the correct answer is B.

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