In Rutherford's model if the electron of Hydrogen atom is going with a velocity v around the nucleus, then find the magnitude of potential energy of the electron.

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Solution

The correct option is B
Potential energy is $(\frac{K q_{1} q_{2}}{r})$ but we don’t know ‘r’ of electron but the velocity v.
But we can find r from v, can’t we?

Electron is going around the nucleus in circular orbit.
So $\frac{m v^{2}}{/ r} = \frac{K q_{1} q_{2}}{r^{/ 2}}$
$r = \frac{K q_{1} q_{2}}{m v^{2}}$
Potential energy is $\frac{K q_{1} q_{2}}{r}$
$p = \frac{K q_{1} q_{2}}{K q_{1} q_{2}} m v^{2}$
$P = m v^{2}$
$P = m_{e} v^{2}$

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In Rutherford's model if the electron of Hydrogen atom is going with a velocity v around the nucleus, then find the magnitude of potential energy of the electron.

The correct option is B Potential energy is (Kq1q2r) but we don’t know ‘r’ of electron but the velocity v. But we can find r from v, can’t we? Electron is going around the nucleus in circular orbit. So mv2/r=Kq1q2r/2 r=Kq1q2mv2 Potential energy is Kq1q2r p=Kq1q2Kq1q2mv2 P=mv2 P=mev2