
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) \[\dfrac{{k{e^2}}}{2}\left( {\dfrac{{{R_2}}}{{{R_1}^2}} - \dfrac{{{R_1}}}{{{R_2}^2}}} \right)\]
Answer
144.9k+ 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.
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
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Physics Average Value and RMS Value JEE Main 2025

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11
