Answer

Verified

35.7k+ views

**Hint:**We need to find the decrease or the change in total energy of the atom. For that we will find the change in kinetic energy of the electron and the change in potential energy separately and then add them to get the change in total energy of the atom.

**Complete step by step solution**

Firstly, we are given that the force of attraction between a positively charged nucleus and the negatively charged electron is $f = k\dfrac{{{e^2}}}{{{r^2}}}$. In the hydrogen atom, this force of attraction will be balanced by the centrifugal force acting on the revolving electron.

\[

\Rightarrow {F_{centripetal}} = f \\

\Rightarrow {F_{centripetal}} = \dfrac{{k{e^2}}}{{{r^2}}} \\

\Rightarrow \dfrac{{m{v^2}}}{r} = \dfrac{{k{e^2}}}{{{r^2}}} \\

\]

Now we will separate out $m{v^2}$ on one side and rest of the terms on the other side.

\[

\Rightarrow m{v^2} = \dfrac{{k{e^2}}}{{{r^2}}} \times r \\

\Rightarrow m{v^2} = \dfrac{{k{e^2}}}{r} \\

\]

Now we know that the kinetic energy for the electron will be${E_{kinetic}} = \dfrac{1}{2}m{v^2}$.

$ \Rightarrow {E_{kinetic}} = \dfrac{1}{2}\dfrac{{k{e^2}}}{r}$

The electron was moving initially in an orbit of radius $R_1$ and later jumped into an orbit of radius $R_2$. The change in kinetic energy between the initial and final states will be

$

\Rightarrow \Delta {E_{kinetic}} = \dfrac{1}{2}\left( {\dfrac{{k{e^2}}}{{{R_2}}} - \dfrac{{k{e^2}}}{{{R_1}}}} \right) \\

\Rightarrow \Delta {E_{kinetic}} = \dfrac{{k{e^2}}}{2}\left( {\dfrac{1}{{{R_2}}} - \dfrac{1}{{{R_1}}}} \right) \\

$

For potential energy, $\Delta {E_{potential}} = - \int\limits_{{R_1}}^{{R_2}} {f.dr} $,where we are given the expression for the force of attraction f.

\[

\Rightarrow \Delta {E_{potential}} = - \int\limits_{{R_1}}^{{R_2}} {\dfrac{{k{e^2}}}{{{r^2}}}.dr} \\

\Rightarrow \Delta {E_{potential}} = - k{e^2}\int\limits_{{R_1}}^{{R_2}} {\dfrac{1}{{{r^2}}}.dr} \\

\Rightarrow \Delta {E_{potential}} = - k{e^2}\left( { - \dfrac{1}{r}} \right)_{{R_1}}^{{R_2}} \\

\Rightarrow \Delta {E_{potential}} = - k{e^2}\left( {\dfrac{1}{{{R_2}}} - \dfrac{1}{{{R_1}}}} \right) \\

\]

So, the total energy will be $\Delta {E_{total}} = \Delta {E_{kinetic}} + \Delta {E_{potential}}$. Substituting the obtained expression for the change in kinetic and potential energy

\[

\Rightarrow \Delta {E_{total}} = \dfrac{{k{e^2}}}{2}\left( {\dfrac{1}{{{R_2}}} - \dfrac{1}{{{R_1}}}} \right) - k{e^2}\left( {\dfrac{1}{{{R_2}}} - \dfrac{1}{{{R_1}}}} \right) \\

\Rightarrow \Delta {E_{total}} = - \dfrac{{k{e^2}}}{2}\left( {\dfrac{1}{{{R_2}}} - \dfrac{1}{{{R_1}}}} \right) \\

or\Delta {E_{total}} = \dfrac{{k{e^2}}}{2}\left( {\dfrac{1}{{{R_2}}} - \dfrac{1}{{{R_1}}}} \right) \\

\]

**Therefore, Option (C) is correct.**

**Note:**We began the solution by equating the centrifugal force and the force of attraction between electron and nucleus. The significance of this is that the electron is negatively charged and the nucleus is positively charged, so the nucleus exerts a force of attraction on the electron, this should mean that the electron falls into the nucleus and sticks to it. But this does not happen because the attractive force on the electron towards the nucleus is counterbalanced by the outwards centrifugal force acting on the electron due to revolution around the nucleus in a curved orbit. Had they not been equal, then either the electron would have ruptured into the nucleus or the electron would have escaped its orbit, leaving the atom unstable in either of the cases. So, the argument used in equating both forces is the cause of stability of an atom.

Recently Updated Pages

To get a maximum current in an external resistance class 1 physics JEE_Main

f a body travels with constant acceleration which of class 1 physics JEE_Main

A hollow sphere of mass M and radius R is rotating class 1 physics JEE_Main

If the beams of electrons and protons move parallel class 1 physics JEE_Main

Two radioactive nuclei P and Q in a given sample decay class 1 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Other Pages

Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main

A gas is compressed isothermally to half its initial class 11 physics JEE_Main

In a CO molecule the distance between C mass12 amu class 12 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main