Consider a neutron and an electron bound to each other due to gravitational force. Assuming Bohr's quantization rule for angular momentum to be vaild in this case, derive an expression for the energy of the neutron -electron system.

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Solution

$M_{e} v r = \frac{n h}{2 π}$ ..........(1)

$\frac{G m_{n} m_{e}}{r^{2}} = \frac{m_{e} v^{2}}{r}$

$\Rightarrow \frac{G m_{n}}{r} = v^{2}$ .......(2)

Squaring (2) and dividing it with (1)

$\frac{m_{e}^{2} v^{2} r^{2}}{v^{2}} = \frac{n^{2} h^{2} r}{4 π^{2} G m_{n}}$

$\Rightarrow m_{e}^{2} r = \frac{n^{2} h^{2} r}{4 π^{2} G m_{n}}$

$\Rightarrow r = \frac{n^{2} h^{2}}{4 π^{2} G m_{n} m_{e}^{2}}$

$\Rightarrow v = \frac{n h}{2 π r m_{e}}$

$\Rightarrow K E = \frac{1}{2} m_{0} v^{2}$

= $\frac{1}{2} m_{0} \frac{(2 π G m_{n} m_{e})^{2}}{n h}$

= $\frac{4 π^{2} G^{2} m_{n}^{2} m_{e}^{3}}{2 n^{2} h^{2}}$

PE = $\frac{- G m_{n} m_{e}}{r}$

= $\frac{- 4 π^{2} G^{2} m_{n}^{2} m_{e}^{2}}{n n^{2} h^{2}}$

= $\frac{- 4 π^{2} G^{2} m_{n}^{2} m_{e}^{2}}{2 n^{2} h^{2}}$

Total energy = KE + PE

= $\frac{2 a^{2} G^{2} m_{n}^{2} m_{e}^{2}}{n^{2} h^{2}}$

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