Question

An electron in an atom jumps in such a way that its kinetic energy changes from $x$ to $\dfrac{x}{9}$. The change in potential energy will be ?

Answer 1

The energy of an electron decreases when it jumps from a higher orbit to a lower orbit in a given atom. Conversely, the energy of an electron increases when it jumps from lower orbit to higher orbit in a given atom.
The change in energy of an electron when it jumps from one orbit to another orbit in a given atom is equal to the difference between its final energy and its initial energy.

Formula used:
The kinetic energy of an electron in an orbit of a given atom is,
$K.E = \dfrac{{Z{e^2}}}{{8\pi {\varepsilon _0}r}}$
And the potential energy of that electron is,
$P.E = \dfrac{{ - Z{e^2}}}{{4\pi {\varepsilon _0}r}}$
Where $Z =$ total number of protons in the nucleus of the given atom, $e =$ charge of electron, ${\varepsilon _0} =$permittivity of free space and $r =$ distance of electron from the nucleus.

Complete step by step answer:
It is given that the initial kinetic energy ${\left( {K.E} \right)_i}$ of the electron is $x$.
${\left( {K.E} \right)_i} = \,\dfrac{{Z{e^2}}}{{8\pi {\varepsilon _0}{r_i}}}$
$\Rightarrow {\left( {K.E} \right)_i} = x$
Where, ${r_i}$ is the radius of the initial orbit of an electron.
And the final kinetic energy ${\left( {K.E} \right)_f}$ of the electron is $\dfrac{x}{9}$.
${\left( {K.E} \right)_f} = \dfrac{{Z{r^2}}}{{8\pi {\varepsilon _0}{r_f}}}$
$\Rightarrow {\left( {K.E} \right)_f} = \dfrac{x}{9}$
Where, ${r_f}$ is the radius of the final orbit of an electron.
The change in kinetic energy $\Delta K.E$ of the electron is given by
${\left( {K.E} \right)_f} - {\left( {K.E} \right)_i} = \dfrac{{Z{r^2}}}{{8\pi {\varepsilon _0}{r_f}}} - \dfrac{{Z{r^2}}}{{8\pi {\varepsilon _0}{r_i}}}$

Substitute the values of ${\left( {K.E} \right)_f}$ and ${\left( {K.E} \right)_i}$ in the above equation.
$\Rightarrow \dfrac{x}{9} - x = \dfrac{{Z{r^2}}}{{8\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r_f}}} - \dfrac{1}{{{r_i}}}} \right)$
Further simplifying
$\Rightarrow - \dfrac{{8x}}{9} = \dfrac{{Z{r^2}}}{{8\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r_f}}} - \dfrac{1}{{{r_i}}}} \right)$
Or $\dfrac{{Z{r^2}}}{{8\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r_f}}} - \dfrac{1}{{{r_i}}}} \right) = - \dfrac{{8x}}{9}$
Let the initial potential energy of the electron is ${\left( {P.E} \right)_i} = - \dfrac{{Z{r^2}}}{{4\pi {\varepsilon _0}{r_i}}}$.
And the final potential energy of the electron is ${\left( {P.E} \right)_f} = - \dfrac{{Z{e^2}}}{{4\pi {\varepsilon _0}{r_f}}}$.

The change in potential energy of the electron $\Delta \left( {P.E} \right)$ is
$\Delta \left( {P.E} \right) = {\left( {P.E} \right)_f} - {\left( {P.E} \right)_i}$
Substituting the value of ${\left( {P.E} \right)_f}$ and ${\left( {P.E} \right)_i}$ in the above formula.
$\Delta \left( {P.E} \right) = - \dfrac{{Z{r^2}}}{{4\pi {\varepsilon _0}{r_f}}} - \left( { - \dfrac{{Z{r^2}}}{{4\pi {\varepsilon _0}{r_i}}}} \right)$
Further simplifying
$\Delta \left( {P.E} \right) = - \dfrac{{Z{r^2}}}{{4\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r_f}}} - \dfrac{1}{{{r_i}}}} \right)$

Now multiply $2$ on the numerator and denominator of the left side of the above equation.
$\Delta \left( {P.E} \right) = - 2\left( {\dfrac{{Z{r^2}}}{{8\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r_f}}} - \dfrac{1}{{{r_i}}}} \right)} \right)$
But we obtained that $\dfrac{{Z{r^2}}}{{8\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{r_f}}} - \dfrac{1}{{{r_i}}}} \right) = - \dfrac{{8x}}{9}$
$\Delta \left( {P.E} \right) = - 2\left( { - \dfrac{{8x}}{9}} \right)$
Further simplifying
$\therefore \Delta \left( {P.E} \right) = \dfrac{{16x}}{9}$

Hence, the change in potential energy of the electron is $\dfrac{{16x}}{9}$.

Note: Alternative method: The total energy of an electron is equal to the sum of kinetic energy and potential energy.
$E = K.E + P.E$
Also, the total energy of an electron is equal to the negative of kinetic energy.
$E = - \left( {K.E} \right)$
So, $K.E + P.E = - \left( {K.E} \right)$
$P.E = - 2\left( {K.E} \right)$
$\Rightarrow \Delta \left( {P.E} \right) = - 2\Delta \left( {K.E} \right)$
It is given that the initial kinetic energy ${\left( {K.E} \right)_i} = x$.
The final kinetic energy ${\left( {K.E} \right)_f} = \dfrac{x}{9}$.
The change in kinetic energy $\Delta \left( {K.E} \right) = {\left( {K.E} \right)_f} - {\left( {K.E} \right)_i}$
Substituting the required values in the above formula.
$\Delta \left( {K.E} \right) = x - \dfrac{x}{9}$
$\Rightarrow \Delta \left( {K.E} \right) = - \dfrac{{8x}}{9}$
But we know that $\Delta \left( {P.E} \right) = - 2\Delta \left( {K.E} \right)$
Substitute the value of $\Delta \left( {K.E} \right)$ in the above formula.
$\Delta \left( {P.E} \right) = - 2\left( { - \dfrac{{8x}}{9}} \right)$
On further simplification
$\Delta \left( {P.E} \right) = \dfrac{{16x}}{9}$
Hence, the change in potential energy of the electron is $\dfrac{{16x}}{9}$.