Answer:
2.068 x 10^6 m / s
Explanation:
radius, r = 5.92 x 10^-11 m
mass of electron, m = 9.1 x 10^-31 kg
charge of electron, q = 1.6 x 10^-19 C
As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.
centripetal force = [tex]\frac{mv^{2}}{r}[/tex]
Electrostatic force = [tex]\frac{kq^{2}}{r^{2}}[/tex]
where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2
So, balancing both the forces we get
[tex]\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}[/tex]
[tex]v=\sqrt{\frac{kq^{2}}{mr}}[/tex]
[tex]v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}[/tex]
v = 2.068 x 10^6 m / s
Thus, the speed of the electron is give by 2.068 x 10^6 m / s.
