Positronium is a system consisting of an electron and its antiparticle a positron, bound together by their electrostatic attraction. Assume that both particles revolve about their common centre of mass. Mark the CORRECT options remaining in the framework of Bohr’s hydrogen atom model. Use $K_{e} - \frac{1}{4 π ε_{0}}, h = \frac{h}{2 π}$ and standard notations.

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Radius of n^th orbit is $r_{n} = (\frac{ℏ^{2}}{m k_{e}^{2} e^{2}}) n^{2}$

Speed in n^th orbit is $u_{n} = (\frac{K_{e} e^{2}}{2 ℏ}) \frac{1}{n}$

Total energy of the bound state is $E_{n} = - \frac{m {(k_{e} e^{2})}^{2}}{4 ℏ^{2}} \frac{1}{n^{2}}$

The magnitudes of the energy levels are exactly one-half of the corresponding value of Bohr’s H-atom.

answer is B, C, D.

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Detailed Solution

Force Equation

$\frac{k e^{2}}{{(2 r_{n})}^{2}} = \frac{m v_{n}^{2}}{r_{n}} \dots \dots \dots \dots \dots \dots \dots (1)$

From Quantization of angular momentum

$m v_{n} r_{n} + m v_{n} r_{n} = \frac{n h}{2 π} \Rightarrow 2 m v_{n} r_{n} = \frac{n h}{2 π} \dots \dots \dots \dots \dots \dots (2)$

Solving equation (1) & (2) we get

$\begin{array}{l} r_{n} = (\frac{ℏ^{2}}{m k e^{2}}) n^{2} \\ v_{n} = (\frac{k e^{2}}{2 h}) \frac{1}{n} \end{array}$

Total Energy $E_{n} = \frac{{(P E)}_{n}}{2}$

$E_{n} = \frac{- k e^{2}}{2 (2 r_{n})}$

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