Atoms & Nuclei

JEE Main level, single correct, medium

82 of 239

Question 82 of 239

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{n}^{th} orbit is proportional to: (n=n= principal quantum number)

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Answer: (D) n−5{n}^{-5}

v∝znv\propto \dfrac{z}{n}

r∝n2zr\propto \dfrac{{n}^{2}}{z}

B=μI2RB=\dfrac{\mu I}{2R}

I=eT=e(2πRv)=e2πvRI=\dfrac{e}{T}=\dfrac{e}{(\dfrac{2\pi R}{v})}=\dfrac{e}{2\pi }\dfrac{v}{R}

B=μ02e2πvR2B=\dfrac{{\mu }_{0}}{2}\dfrac{e}{2\pi }\dfrac{v}{{R}^{2}}

B=μ0e4πzn(n2z)2B=\dfrac{{\mu }_{0}e}{4\pi }\dfrac{z}{n{(\dfrac{{n}^{2}}{z})}^{2}}

B=μ0e4πz3n5B=\dfrac{{\mu }_{0}e}{4\pi }\dfrac{{z}^{3}}{{n}^{5}}

B∝n−5B\propto {n}^{-5}.

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