Verified Solution • 2025–26 Session moderate

QJEEM17Ph3886: JEE Main 2017 Physics - Class 12 JEEM17Ph3886 Solution: According to Bohr’s theory, the time averaged magnetic field at …

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)

Multiple Choice Options

  • A n−4{n^{ - 4}}
  • B n−5{n^{ - 5}}
  • C n −- 3
  • D n −- 2

Step-by-Step Verified Solution

Magnetic field at the center of neucleus of H-atom

B = μ0I2rn{{{\mu _0}I} \over {2{r_n}}}

Radius of nth orbital,

rn = n2h2ε0mπZe2{{{n^2}{h^2}{\varepsilon _0}} \over {m\pi Z{e^2}}}

∴   \therefore\,\,\, rn ∝\propto n2

velocity of electron in nth orbital,

υ\upsilon n = (e2h2ε0)Zn\left( {{{{e^2}h} \over {2{\varepsilon _0}}}} \right){Z \over n}

∴   \therefore\,\,\, υ\upsilon n ∝\propto n −- 1

I = qt{q \over t} = e2πrnυn{e \over {{{2\pi {r_n}} \over {{\upsilon _n}}}}} = eυn2πrn{{e{\upsilon _n}} \over {2\pi {r_n}}}

∴   \therefore\,\,\, B = μ0.(eυn2πrn)2rn{{{\mu _0}.\left( {{{e{\upsilon _n}} \over {2\pi {r_n}}}} \right)} \over {2{r_n}}} = μ0 eυn4πrn2{{{\mu _0}\,e{\upsilon _n}} \over {4\pi r_n^2}}

∴   \therefore\,\,\, B ∝\propto υnrn2{{{\upsilon _n}} \over {r_n^2}}

⇒\Rightarrow    \,\,\, B ∝\propto n−1n4{{{n^{ - 1}}} \over {{n^4}}}

⇒\Rightarrow    \,\,\, B ∝\propto n −- 5

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