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Kinetic Energy

The force of ...

Question

The force of attraction between the positively charged nucleus and the electron in a hydrogen atom is given by $f = k \frac{e^{2}}{r^{2}}$ . Assume that the nucleus is fixed. The electron, initially moving in an orbit of radius $R_{1}$ jumps into an orbit of smaller radius $R_{2}$ . the decreases in the total energy of the atom is.

\frac{k e^{2}}{2} (\frac{1}{R_{1}} - \frac{1}{R_{2}})

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\frac{k e^{2}}{2} (\frac{R_{1}}{R_{2}^{2}} - \frac{R_{2}}{R_{1}^{2}})

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\frac{k e^{2}}{2} (\frac{1}{R_{2}} - \frac{1}{R_{1}})

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\frac{k e^{2}}{2} (\frac{R_{2}}{R_{1}^{2}} - \frac{R}{R_{2}^{2}})

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Solution

The correct option is A $\frac{k e^{2}}{2} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
It is given that,

$f = \frac{k e^{2}}{r^{2}}$

This force is balanced by the centripetal force.

$f_{c} = \frac{k e^{2}}{r^{2}}$

$\frac{m v^{2}}{r} = \frac{k e^{2}}{r^{2}}$

$m v^{2} = \frac{k e^{2}}{r}$

Kinetic energy of electron is $= \frac{1}{2} m v^{2} = \frac{1}{2} \frac{k e^{2}}{R}$

The electron is moving in a circle of radius R₁ and jumps into the circle of radius R₂.

So,

$Δ K = \frac{1}{2} k e^{2} [\frac{1}{R_{2}} - \frac{1}{R_{1}}] . . . . . . . . . . . . (1)$

Potential energy is

$Δ U = - R_{2} \int R_{1} f . d r$

$= - R_{2} \int R_{1} \frac{k e^{2}}{r^{2}} d r$

$= - k e^{2} (\frac{1}{R_{2}} - \frac{1}{R_{1}})$

So, change in total energy is

$Δ E = Δ K + Δ U$

$= \frac{1}{2} k e^{2} [\frac{1}{R_{2}} - \frac{1}{R_{1}}] + - k e^{2} (\frac{1}{R_{2}} - \frac{1}{R_{1}})$

$= - \frac{1}{2} k e^{2} (\frac{1}{R_{2}} - \frac{1}{R_{1}})$
or
$Δ E = \frac{1}{2} k e^{2} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

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