
Suppose potential energy between electron and proton at separation \[r\] is given by \[U = k\,ln\left( r \right)\] , where \[k\] is constant. For such hypothetical hydrogen atom, the ratio of energy difference between energy levels of ( \[n = 1\] to \[n = 2\] ) and ( \[n = 2\] to \[n = 4\] ) is
A. \[1\]
B. \[2\]
C. \[\dfrac{1}{2} \\ \]
D. \[3\]
Answer
161.1k+ views
Hint:To solve this question first calculate the expression of electrostatic force from the expression of the potential energy given in the question. This force is balanced by the centripetal force on the electron revolving around the nucleus. Also, the total energy is sum of the kinetic energy and the potential energy. Combine these concepts and derive the expression of total energy to find the required ratio.
Formula Used:
Electrostatic force,
$F = - \dfrac{{dU}}{{dr}}$
where $U$ is the potential energy and $r$ is the distance between the two charges.
Centripetal force,
${F_c} = \dfrac{{m{v^2}}}{r}$
where $v$ is the velocity of the charge (here, electron), $m$ is the mass of the object (here, electron) and $r$ is the radius of the orbit (here, distance between electron and proton).
Total energy,
$E = K + U$ where $K$ is the kinetic energy and is equal to $\dfrac{1}{2}m{v^2}$.
Angular momentum,
\[mvr = \dfrac{{nh}}{{2\pi }}\]
Complete step by step solution:
Given: \[U = k\,ln\left( r \right)\]...(1)
We know that, electrostatic force is $F = - \dfrac{{dU}}{{dr}}$
Therefore, differentiating equation (1), with respect to $r$ , we get,
$F = - \dfrac{k}{r}$ ...(2)
This force is balanced by the centripetal force on the electron revolving around the nucleus. This implies that,
$\dfrac{k}{r} = \dfrac{{m{v^2}}}{r}$ ...(3)
Also, angular momentum is given by,
\[mvr = \dfrac{{nh}}{{2\pi }}\] ...(4)
Also, $E = K + U$ ...(5)
Where $K = \dfrac{1}{2}m{v^2}$ ...(6)
Substituting equations (1) and (6) in equation (5), we get,
$E = \dfrac{1}{2}m{v^2} + k\,\ln \left( r \right) \\ $ ...(7)
Substitute the value of $m{v^2}$ from equation (3) into equation (7), we get,
$E = \dfrac{k}{2} + k\,\ln \left( r \right)$
Now, substitute the value of $r$ from equation (4), we get
$E = \dfrac{k}{2} + \dfrac{k}{2} \times 2\,\ln \left( {\dfrac{{nh}}{{2\pi mv}}} \right) \\ $
Simplifying this,
$E = \dfrac{k}{2}\left( {1 + \ln {{\left( {\dfrac{{nh}}{{2\pi mv}}} \right)}^2}} \right) \\ $
This implies,
$E = \dfrac{k}{2}\left( {1 + \ln \left( {\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right)} \right) \\ $
Now, the energy difference between levels $1$ and $2$ is,
${E_2} - {E_1} = \ln \left( {\dfrac{{4{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right) - \ln \left( {\dfrac{{{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right) \\ $...(8)
Similarly, the energy difference between levels $2$ and $4$ is,
${E_4} - {E_2} = \ln \left( {\dfrac{{16{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right) - \ln \left( {\dfrac{{4{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right)$...(9)
Dividing equation (8) and (9), we get,
\[\dfrac{{{E_2} - {E_1}}}{{{E_4} - {E_2}}} = \dfrac{{\ln \left( {\dfrac{{4{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right) - \ln \left( {\dfrac{{{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right)}}{{\ln \left( {\dfrac{{16{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right) - \ln \left( {\dfrac{{4{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right)}} \\ \]
Solving this, we get,
\[\dfrac{{{E_2} - {E_1}}}{{{E_4} - {E_2}}} = \dfrac{{\ln \left( 4 \right) - \ln \left( 1 \right)}}{{\ln \left( {16} \right) - \ln \left( 4 \right)}} \\ \]
Solving further,
\[\dfrac{{{E_2} - {E_1}}}{{{E_4} - {E_2}}} = \dfrac{{\ln \left( {\dfrac{4}{1}} \right)}}{{\ln \left( {\dfrac{{16}}{4}} \right)}} \\ \]
\[\therefore \dfrac{{{E_2} - {E_1}}}{{{E_4} - {E_2}}} = 1\]
Hence, option A is the answer.
Note: One can make mistakes while substituting one equation into the other, so the equations must be handled carefully. Also, one must know the properties of natural logarithm as it makes the calculations easier and the equations can be simplified. There is no need to put any numerical value in any equation, like the value of Planck’s constant etc.
Formula Used:
Electrostatic force,
$F = - \dfrac{{dU}}{{dr}}$
where $U$ is the potential energy and $r$ is the distance between the two charges.
Centripetal force,
${F_c} = \dfrac{{m{v^2}}}{r}$
where $v$ is the velocity of the charge (here, electron), $m$ is the mass of the object (here, electron) and $r$ is the radius of the orbit (here, distance between electron and proton).
Total energy,
$E = K + U$ where $K$ is the kinetic energy and is equal to $\dfrac{1}{2}m{v^2}$.
Angular momentum,
\[mvr = \dfrac{{nh}}{{2\pi }}\]
Complete step by step solution:
Given: \[U = k\,ln\left( r \right)\]...(1)
We know that, electrostatic force is $F = - \dfrac{{dU}}{{dr}}$
Therefore, differentiating equation (1), with respect to $r$ , we get,
$F = - \dfrac{k}{r}$ ...(2)
This force is balanced by the centripetal force on the electron revolving around the nucleus. This implies that,
$\dfrac{k}{r} = \dfrac{{m{v^2}}}{r}$ ...(3)
Also, angular momentum is given by,
\[mvr = \dfrac{{nh}}{{2\pi }}\] ...(4)
Also, $E = K + U$ ...(5)
Where $K = \dfrac{1}{2}m{v^2}$ ...(6)
Substituting equations (1) and (6) in equation (5), we get,
$E = \dfrac{1}{2}m{v^2} + k\,\ln \left( r \right) \\ $ ...(7)
Substitute the value of $m{v^2}$ from equation (3) into equation (7), we get,
$E = \dfrac{k}{2} + k\,\ln \left( r \right)$
Now, substitute the value of $r$ from equation (4), we get
$E = \dfrac{k}{2} + \dfrac{k}{2} \times 2\,\ln \left( {\dfrac{{nh}}{{2\pi mv}}} \right) \\ $
Simplifying this,
$E = \dfrac{k}{2}\left( {1 + \ln {{\left( {\dfrac{{nh}}{{2\pi mv}}} \right)}^2}} \right) \\ $
This implies,
$E = \dfrac{k}{2}\left( {1 + \ln \left( {\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right)} \right) \\ $
Now, the energy difference between levels $1$ and $2$ is,
${E_2} - {E_1} = \ln \left( {\dfrac{{4{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right) - \ln \left( {\dfrac{{{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right) \\ $...(8)
Similarly, the energy difference between levels $2$ and $4$ is,
${E_4} - {E_2} = \ln \left( {\dfrac{{16{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right) - \ln \left( {\dfrac{{4{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right)$...(9)
Dividing equation (8) and (9), we get,
\[\dfrac{{{E_2} - {E_1}}}{{{E_4} - {E_2}}} = \dfrac{{\ln \left( {\dfrac{{4{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right) - \ln \left( {\dfrac{{{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right)}}{{\ln \left( {\dfrac{{16{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right) - \ln \left( {\dfrac{{4{h^2}}}{{4{\pi ^2}{m^2}{v^2}}}} \right)}} \\ \]
Solving this, we get,
\[\dfrac{{{E_2} - {E_1}}}{{{E_4} - {E_2}}} = \dfrac{{\ln \left( 4 \right) - \ln \left( 1 \right)}}{{\ln \left( {16} \right) - \ln \left( 4 \right)}} \\ \]
Solving further,
\[\dfrac{{{E_2} - {E_1}}}{{{E_4} - {E_2}}} = \dfrac{{\ln \left( {\dfrac{4}{1}} \right)}}{{\ln \left( {\dfrac{{16}}{4}} \right)}} \\ \]
\[\therefore \dfrac{{{E_2} - {E_1}}}{{{E_4} - {E_2}}} = 1\]
Hence, option A is the answer.
Note: One can make mistakes while substituting one equation into the other, so the equations must be handled carefully. Also, one must know the properties of natural logarithm as it makes the calculations easier and the equations can be simplified. There is no need to put any numerical value in any equation, like the value of Planck’s constant etc.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Young's Double Slit Experiment Step by Step Derivation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

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

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

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

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

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