Magnetic ﬁeld at the centre (at nucleus) of the hydrogen like atoms (atomic number = Z) due to the motion of electron in nth orbit is proportional to

\frac{n^{3}}{Z^{5}}

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\frac{n^{4}}{Z}

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\frac{Z^{2}}{n^{3}}

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\frac{Z^{3}}{n^{5}}

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Solution

The correct option is D $\frac{Z^{3}}{n^{5}}$
$B_{n} = \frac{μ_{0} I_{n}}{2 r_{n}} o r B_{n} \propto \frac{I_{n}}{r_{n}} \propto \frac{(f_{n})}{r_{n}} ∴ B_{n} \propto \frac{(\frac{V_{n}}{r_{n}})}{r_{n}} \propto \frac{V_{n}}{(r_{n})^{2}} \propto \frac{(\frac{Z}{n})}{(\frac{n^{2}}{Z^{2}})} \propto \frac{Z^{3}}{n^{5}}$

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Magnetic ﬁeld at the centre (at nucleus) of the hydrogen like atoms (atomic number = Z) due to the motion of electron in nth orbit is proportional to

The correct option is D Z3n5Bn=μ0In2rn or Bn∝Inrn∝(fn)rn∴ Bn∝(Vnrn)rn∝Vn(rn)2∝(Zn)(n2Z2)∝Z3n5