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An electron of mass 9.11x10^-31 is revolving in a stable orbit of radius 5.37×10-34m. If electrostatic force of attraction between electron and proton is 8×10-8N. Find its velocity

An electron of mass 9.11x10^-31 is revolving in a stable orbit of radius 5.37×10-34m. If electrostatic force of attraction between electron and proton is 8×10-8N. Find its velocity

Grade:12

1 Answers

Eshan
askIITians Faculty 2095 Points
5 years ago
Dear student,

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

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