In a hydrogen atom, magnetic field produced by the electron at the centre of circular orbit is ‘B’ when the atom is in ground state. What will be the magnetic field produced by the electron at the centre of circular orbit when the atom is in second excital state?

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$\frac{B}{32}$

$\frac{B}{81}$

$\frac{B}{64}$

$\frac{B}{243}$

answer is C.

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Detailed Solution

$i_{n} = \frac{{eω}_{n}}{2 π} = \frac{{eV}_{n}}{2 {πr}_{n}}; B_{n} = \frac{μ_{o} i_{n}}{2 r_{n}} = \frac{μ_{o} e}{4 π} . \frac{V_{n}}{r_{n}^{2}}$

$Now, V_{n} \propto \frac{1}{n} and r_{n} \propto n^{2}$

$\begin{array}{l} ∴ B_{n} \propto \frac{1}{n^{5}} \\ ∴ \frac{B_{3}}{B} = \frac{1}{{(3)}^{5}} \Rightarrow B_{3} = \frac{B}{243} \end{array}$

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