The KE of the electron in an orbit of radius 'r' in hydrogen atom is proportional to
(e $=$ electronic charge)

\frac{e^{2}}{r^{2}}

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\frac{e^{2}}{2 r^{3}}

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\frac{e^{2}}{r}

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\frac{e^{2}}{2 r^{2}}

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Solution

The correct option is C $\frac{e^{2}}{r}$
From, $\frac{m v^{2}}{r} = \frac{K e^{2}}{r^{2}}$
We get, $m v^{2} = \frac{K e^{2}}{r}$
Thus kinetic energy, $K = \frac{m v^{2}}{2} = \frac{K e^{2}}{2 r}$
$⟹$ $K \propto \frac{e^{2}}{r}$

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The KE of the electron in an orbit of radius 'r' in hydrogen atom is proportional to (e=electronic charge)

The correct option is C e2rFrom, mv2r=Ke2r2 We get, mv2=Ke2rThus kinetic energy, K=mv22=Ke22r⟹ K∝e2r

The KE of the electron in an orbit of radius 'r' in hydrogen atom is proportional to
(e $=$ electronic charge)

The correct option is C $\frac{e^{2}}{r}$
From, $\frac{m v^{2}}{r} = \frac{K e^{2}}{r^{2}}$
We get, $m v^{2} = \frac{K e^{2}}{r}$
Thus kinetic energy, $K = \frac{m v^{2}}{2} = \frac{K e^{2}}{2 r}$
$⟹$ $K \propto \frac{e^{2}}{r}$