The magnitude of total energy and angular mom | vedclass

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(C) According to Bohr's model,the energy of an electron in the $n^{th}$ orbit is given by $E_{n} = -\frac{13.6}{n^{2}} 	ext{ eV}$.Thus,$E_{n} \propto

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$E_{n} \propto \frac{1}{L_{n}^{2}}$

Vedclass · Answer

(C) According to Bohr's model,the energy of an electron in the $n^{th}$ orbit is given by $E_{n} = -\frac{13.6}{n^{2}} 	ext{ eV}$.Thus,$E_{n} \propto \frac{1}{n^{2}}$.The angular momentum of an electron in the $n^{th}$ orbit is given by $L_{n} = \frac{nh}{2\pi}$.Thus,$L_{n} \propto n$,which implies $n \propto L_{n}$.Substituting $n \propto L_{n}$ into the energy relation,we get $E_{n} \propto \frac{1}{L_{n}^{2}}$.Therefore,the correct relation is $E_{n} \propto \frac{1}{L_{n}^{2}}$.

The magnitude of total energy and angular momentum of an electron in the $n^{th}$ orbit of a Bohr atom is denoted by $E_{n}$ and $L_{n}$ respectively. Then

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