Answer:
Electron's centripetal acceleration, [tex]a=5.66\times 10^{23}\ m/s^2[/tex]
Explanation:
Given that,
The radius of the orbit, [tex]r=2.99\times 10^{-11}\ m[/tex]
The centripetal force is balanced by the electrostatic force such that,
[tex]k\dfrac{q_1q_2}{r^2}=\dfrac{mv^2}{r}[/tex]
[tex]a=\dfrac{v^2}{r}=\dfrac{kq_1q_2}{mr^2}[/tex]
Here, [tex]q_1=e[/tex] and [tex]q_2=2e[/tex]
[tex]a=\dfrac{9\times 10^9\times 1.6\times 10^{-19}\times 2\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times (2.99\times 10^{-11})^2}[/tex]
[tex]a=5.66\times 10^{23}\ m/s^2[/tex]
So, the magnitude of the electron's centripetal acceleration [tex]5.66\times 10^{23}\ m/s^2[/tex]. Hence, this is the required solution.
