Respuesta :
Explanation:
According to the energy conservation,
[tex]F_{centripetal} = F_{electric}[/tex]
[tex]\frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}[/tex]
[tex]v^{2} = \frac{kq^{2}r}{d^{2}m}[/tex]
= [tex]\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}[/tex]
= [tex]8.430 \times 10^{10} m^{2}/s^{2}[/tex]
v = [tex]\sqrt{8.430 \times 10^{10} m^{2}/s^{2}}[/tex]
= [tex]2.903 \times 10^{5} m/s[/tex]
Formula for distance from the orbit is as follows.
S = [tex]2 \pi r[/tex]
= [tex]2 \times 3.14 \times 0.75 \times 10^{-9} m[/tex]
= [tex]4.71 \times 10^{-9} m[/tex]
Now, relation between time and distance is as follows.
T = [tex]\frac{S}{v}[/tex]
[tex]\frac{1}{f} = \frac{S}{v}[/tex]
or, f = [tex]\frac{v}{S}[/tex]
= [tex]\frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}[/tex]
= [tex]6.164 \times 10^{13} Hz[/tex]
Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is [tex]6.164 \times 10^{13} Hz[/tex].
