Question

A positron is an elementary particle identical to an electron except that its charge is $+e$. An electron and a positron can rotate about their center of mass as if they were a dumbbell connected by a massless rod. What is the orbital frequency for an electron and a positron$1.0nm$ apart?

Short answer

The orbital frequency for the electron and positron is$\mathbf{1}\mathbf{.}\mathbf{125}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}\mathbf{}}\mathbf{Hz}$

Step-by-step solution

Given information and formula used

Given :

Distance between the electron and positron : $1.0nm$

Theory used :

The centripetal force is generated by electrostatic force. That is,

$$\begin{gathered} F_e=F_c \\ \Rightarrow \frac{1}{4\pi {\epsilon }_0}\times \frac{e^2}{d^2}=mw²\frac{d}{2} \end{gathered}$$

Evaluating the required orbital frequency 

The positron has the same mass as the electron, and the distance between them is $d=1x{10}^{-9}$.

The mass is equal, and the center of mass is at role="math" localid="1649128476171" $\frac{d}{2}=0.5x{10}^{-9}$

The centripetal force is generated by electrostatic force. So

$$\begin{gathered} F_e=F_c \\ \Rightarrow \frac{1}{4\pi {\epsilon }_0}\times \frac{e^2}{d^2}=mw²\frac{d}{2} \\ \Rightarrow w^2=\frac{9\times {10}^{-9}\times {(1.6x{10}^{-19})}^2\times 2}{(9.11\times {10}^{-31}){(1x{10}^{-9})}^3} \\ \Rightarrow w=\underline{7x{10}^{5}rad/s} \end{gathered}$$

We now have orbital frequency :

$f=\frac{w}{2\pi }=\frac{7x{10}^5}{2\times 3.14}=\boxed{\mathbf{1}\mathbf{.}\mathbf{125}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{H}\mathit{z}\mathbf{}}$