Question

A positronium consists of an electron and a positron, revolving about their common centre of mass. Calculate the kinetic energy of the electron if the radius of circular path is equal to the radius of electron in ground state of a hydrogen atom.

• A. $1.51\text{eV}$
• B. $3.4\text{eV}$
• C. $6.8\text{eV}$
• D. $13.6\text{eV}$

Answer 1

The correct option is B $3.4\text{eV}$

Let $r$ be the distance between electron and positron and $r^{\prime }$ be the radius of the circular path.

As, electron and positron have same mass,
$\therefore r^{\prime }=\frac{r}{2}$

Here, the centripetal force is provided by the electrostatic force of attraction between the electron and the positron.

$\Rightarrow \frac{k{e}^2}{r^2}=\frac{m{v}^2}{r^{\prime }}$

$\Rightarrow m{v}^2=\frac{k{e}^2{r}^{\prime }}{(2r^{\prime }{)}^2}=\frac{k{e}^2}{4r^{\prime }}$

$K.E=\frac{1}{2}m{v}^2=\frac{k{e}^2}{8r^{\prime }}$

For an electron in the ground state of a hydrogen atom, $r^{\prime }=0.53\dot{A}$

$\Rightarrow K.E=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{8\times \left(0.53\dot{A}\right)}$

$=\frac{13.6\text{eV}}{4}=3.4\text{eV}$

Hence, option $\left(B\right)$ is correct.