If $E$ and $J_{n}$ are the magnitude of total energy and angular momentum of electron in the nth Bohr orbit respectively, then,

E \propto J_{n}^{2}

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E \propto \frac{1}{J_{n}^{2}}

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E \propto J_{n}

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E \propto \frac{1}{J_{n}}

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Solution

The correct option is B $E \propto \frac{1}{J_{n}^{2}}$
$J_{n} = l = n \frac{h}{2 R} = m v r$ - (1) (In Bohr's model)
And from Coulomb force:
$F = \frac{Z e^{2}}{4 n E_{0} r^{2}}$
And from centripetal force:
$\frac{Z e^{2}}{4 n E_{0} r^{2}} = \frac{m V^{2}}{r}$
$r = \frac{Z e^{2}}{4 n E_{0} m V^{2}}$ ------ (2)
Eliminating $r$ from (1) and (2)
$V = \frac{Z e^{2}}{2 E_{0} h n}$
$r = \frac{E_{0} h^{2} n^{2}}{R m Z e^{2}}$
Now, Kinetic energy $= \frac{1}{2} m v^{2} = \frac{m z^{2} e^{4}}{8 E_{0}^{2} h^{2} n^{2}}$
Potential energy $= \frac{- Z e^{2}}{4 R E_{0} r} = \frac{- m Z^{2} e^{4}}{4 E_{0}^{2} h^{2} n^{2}}$
So, total energy $= K + V =$ $\frac{- m Z^{2} e^{4}}{8 E_{0}^{2} h^{2} n^{2}}$ $= \frac{- m Z^{2} e^{4} 4 R^{2}}{8 E_{0}^{2} (4 R^{2}) h^{2} n^{2}}$ $= \frac{- m Z^{2} e^{4}}{32 E_{0}^{2} R^{2}} \times \frac{1}{l^{2}}$

So, $E \propto \frac{1}{{J n}^{2}}$

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