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

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Solution

According to Bohr's quantization rule,

Angular momentum of electron, L = $\frac{n h}{2 π}$
$\Rightarrow m v_{e} r = \frac{n h}{2 π} \Rightarrow v_{e} = \frac{n h}{2 π r m_{e}}$ ...(1)
Here,
n = Quantum number
h = Planck's constant
m = Mass of the electron
r = Radius of the circular orbit
v_e = Velocity of the electron

Let m_n be the mass of neutron.

On equating the gravitational force between neutron and electron with the centripetal acceleration,
$\frac{G m_{n} m_{e}}{r^{2}} = \frac{m_{e} v^{2}}{r} \Rightarrow \frac{G m_{n}}{r} = v^{2} . . . (2)$
Squaring (1) and dividing it by (2), we have
$\frac{m_{e}^{2} v^{2} r^{2}}{v^{2}} = \frac{n^{2} h^{2} r}{4 π^{2} G m_{n}} \Rightarrow m_{e}^{2} r^{2} = \frac{n^{2} h^{2} r}{4 π^{2} G m_{n}} \Rightarrow r = \frac{n^{2} h^{2}}{4 π^{2} G m_{n} m_{e}^{2}} v_{e} = \frac{n h}{2 π r m_{e}} \Rightarrow v_{e} = \frac{n h}{2 π m_{e} \times (\frac{n^{2} h^{2}}{4 π^{2} G m_{n} m_{e}^{2}})} \Rightarrow v_{e} = \frac{2 π G m_{n} m_{e}}{n h}$

$Kinetic e nergy of the electron, K = \frac{1}{2} m_{e} {v_{e}}^{2} = \frac{1}{2} m_{e} {(\frac{2 πG m_{n} m_{e}}{n h})}^{2} = \frac{4 π^{2} G^{2} m_{n}^{2} m_{e}^{3}}{2 n^{2} h^{2}}$

Potential energy of the neutron, $P = \frac{- G m_{n} m_{e}}{r}$

Substituting the value of r in the above expression,
$P = \frac{- G m_{e} m_{n} 4 π^{2} G m_{n} m_{e}^{2}}{n^{2} h^{2}} P = \frac{- 4 π^{2} G^{2} m_{n}^{2} m_{e}^{3}}{n^{2} h^{2}}$
Total energy = K + P $= - \frac{2 π^{2} G^{2} m_{n}^{2} m_{e}^{2}}{2 n^{2} h^{2}} = - \frac{π^{2} G^{2} m_{n}^{2} m_{e}^{2}}{n^{2} h^{2}}$

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