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?

Answer 1

Answer:

$1.1312947443\times 10^{14}\ Hz$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge of electron = $1.6\times 10^{-19}\ C$

d = Distance = 1 nm

v = Velocity of positron

In this system the centripetal and electrical forces are conserved

$\dfrac{mv^2}{r}=\dfrac{kq^2}{d^2}\\\Rightarrow \dfrac{mv^2}{\dfrac{d}{2}}=\dfrac{kq^2}{d^2}\\\Rightarrow v=\sqrt{\dfrac{kq^2}{2md}}\\\Rightarrow v=\sqrt{\dfrac{8.99\times 10^{9}\times (1.6\times 10^{-19})^2}{2\times 9.11\times 10^{-31}\times 1\times 10^{-9}}}\\\Rightarrow v=355406.725773\ m/s$

Distance to orbit is given by

$s=2\pi r\\\Rightarrow s=2\pi 0.5\times 10^{-9}\\\Rightarrow s=3.1415926536\times 10^{-9}\ m$

Time is given by

$t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{3.1415926536\times 10^{-9}}{355406.725773}\\\Rightarrow t=8.8394293799\times 10^{-15}\ s$

Frequency is given by

$f=\dfrac{1}{T}\\\Rightarrow f=\dfrac{1}{8.8394293799\times 10^{-15}}\\\Rightarrow f=1.1312947443\times 10^{14}\ Hz$

The orbital frequency is $1.1312947443\times 10^{14}\ Hz$