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# The force of attraction between the positively charged nucleus and the electron in a hydrogen atom is given by $f = k\dfrac{{{e^2}}}{{{r^2}}}$ . 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 decrease in the total energy of the atom is (A) $k\dfrac{{{e^2}}}{{{r^2}}}\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$(B) $\dfrac{{k{e^2}}}{2}\left( {\dfrac{{{R_1}}}{{{R_2}^2}} - \dfrac{{{R_2}}}{{{R_1}^2}}} \right)$(C) $\dfrac{{k{e^2}}}{2}\left( {\dfrac{1}{{{R_2}}} - \dfrac{1}{{{R_1}}}} \right)$(D) $\dfrac{{k{e^2}}}{2}\left( {\dfrac{{{R_2}}}{{{R_1}^2}} - \dfrac{{{R_1}}}{{{R_2}^2}}} \right)$

Last updated date: 21st Apr 2024
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Answer
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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.