Question

19) An electron of mass 9 x10-31 kg is revolving in
a stable circular orbit of radius 1.8 x 10-12 min
a hydrogen atom. If the electrostatic force of
attraction between the proton and electron is 8
x 10-8 N, the velocity of the electron is​

Answer 1

Dear friend

Force between the nucleus and the electron=\dfrac{1}{4\pi\epsilon_0}\dfrac{Ze.e}{r^2}=8\times 10^{-8}

Force between the nucleus and the electron=\dfrac{1}{4\pi\epsilon_0}\dfrac{Ze.e}{r^2}=8\times 10^{-8}The centripetal acceleration of the electron is provided by the electrostatic attraction between the electron and the proton.

Force between the nucleus and the electron=\dfrac{1}{4\pi\epsilon_0}\dfrac{Ze.e}{r^2}=8\times 10^{-8}The centripetal acceleration of the electron is provided by the electrostatic attraction between the electron and the proton.\implies \dfrac{1}{4\pi\epsilon_0}\dfrac{Ze.e}{r^2}=\dfrac{m_ev^2}{r}=8\times 10^{-8}

Force between the nucleus and the electron=\dfrac{1}{4\pi\epsilon_0}\dfrac{Ze.e}{r^2}=8\times 10^{-8}The centripetal acceleration of the electron is provided by the electrostatic attraction between the electron and the proton.\implies \dfrac{1}{4\pi\epsilon_0}\dfrac{Ze.e}{r^2}=\dfrac{m_ev^2}{r}=8\times 10^{-8}\implies v=\sqrt{\dfrac{8\times 10^{-8}\times 5.37\times 10^{-34}}{9.1\times 10^{-31}}}

Force between the nucleus and the electron=\dfrac{1}{4\pi\epsilon_0}\dfrac{Ze.e}{r^2}=8\times 10^{-8}The centripetal acceleration of the electron is provided by the electrostatic attraction between the electron and the proton.\implies \dfrac{1}{4\pi\epsilon_0}\dfrac{Ze.e}{r^2}=\dfrac{m_ev^2}{r}=8\times 10^{-8}\implies v=\sqrt{\dfrac{8\times 10^{-8}\times 5.37\times 10^{-34}}{9.1\times 10^{-31}}}=6.87\times 10^{-6}m/s.

@Rajsingh...