In Bohr's model of the hydrogen atom:

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The radius of the nth orbit is inversely proportional to $n^{2}$

The total energy of the electron in the nth orbit is inversely proportional to 'n .

The angular momentum of the electron in an orbit is an integral multiple of $h / 2 π$ .

The magnitude of potential energy of the electron in any orbit is greater than its KE.

answer is A, C, D.

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Detailed Solution

The expression of radius of n th orbit for H-atom: $r_{n} = \frac{n^{2} h^{2}}{4 π^{2} e^{2} mk}$ where k = $1 / 4 π ϵ_{0}$ .

So, $r_{n} \propto n^{2}$

The KE of electron for H-atom is $K_{n} = \frac{k e^{2}}{2 r_{n}}$ or potential energy, $U_{n} = - \frac{k e^{2}}{1_{n}}$

Total energy $E = K_{n} + U_{n} = - \frac{{ke}^{2}}{2 r_{n}}$ So $E \propto 1 / r_{n} = 1 / n^{2}$

Also, $|U_{n}| > K_{n}$

From Bohr's second postulate, angular momentum, $L = mvr = nh / 2 π$ .

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