Question

A positron is an elementary particle identical to an electron except that its charge is . 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 1.50 apart?

Answer 1

Explanation:

According to the energy conservation,

          $F_{centripetal} = F_{electric}$

            $\frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}$

           $v^{2} = \frac{kq^{2}r}{d^{2}m}$

                 = $\frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}$

                = $8.430 \times 10^{10} m^{2}/s^{2}$

             v = $\sqrt{8.430 \times 10^{10} m^{2}/s^{2}}$

                = $2.903 \times 10^{5} m/s$

Formula for distance from the orbit is as follows.

               S = $2 \pi r$

                  = $2 \times 3.14 \times 0.75 \times 10^{-9} m$

                  = $4.71 \times 10^{-9} m$

Now, relation between time and distance is as follows.

                T = $\frac{S}{v}$

       $\frac{1}{f} = \frac{S}{v}$

or,           f = $\frac{v}{S}$          

                = $\frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}$      

                = $6.164 \times 10^{13} Hz$

Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is $6.164 \times 10^{13} Hz$.