Question

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 1

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.