Question

The potential energy between the electron and proton is given by $U = K{e^2}/3{r^3}$ . According to Bohr’s theory, the energy in the nth orbit of such a hypothetical atom will be proportional to
(A) ${n^6}$
(B) ${n^4}$
(C) ${n^2}$
(D) $n$

Answer 1

Hint:From the potential energy, calculate the force on the electron by using the formula. Equate the obtained force with that of the centripetal force and find the velocity value. Use the formula of momentum and calculate the value of radius and with the help of that find total energy.

Useful formula:
(1) The formula of the force experienced by the electron is given by
$F = - \dfrac{{dU}}{{dr}}$
Where $F$ is the force on the electron, $r$ is the radius of the orbit and $U$ is the potential energy.
(2) The centripetal force is given by
${F_c} = \dfrac{{m{v^2}}}{r}$
Where ${F_c}$ is the centripetal force, $m$ is the mass of the electron, $v$ is the velocity of the electron.
(3) The formula for the angular momentum is given by
$L = \dfrac{{nh}}{{2\pi }}$
Where $L$ is the angular momentum, $n$ is the orbit of the electron and $h$ is the Planck’s constant.

Complete step by step solution:
It is given that the
The potential energy between the electron and proton, $U = \dfrac{{K{e^2}}}{{3{r^3}}}$
Using the formula of the force,
$F = - \dfrac{{dU}}{{dr}}$
Substituting the equation of the potential energy in it.
$F = - \dfrac{{d\left( {\dfrac{{K{e^2}}}{{3{r^3}}}} \right)}}{{dr}}$
$F = \dfrac{{K{e^2}}}{{{r^4}}}$
The above force is balanced by the centripetal force, hence
$F = {F_c}$
$\dfrac{{m{v^2}}}{r} = \dfrac{{K{e^2}}}{{{r^4}}}$
By further simplification,
${v^2} = \dfrac{{K{e^2}}}{{m{r^3}}}$ ---------------(1)
Using the formula (3)
$L = \dfrac{{nh}}{{2\pi }}$
Substituting the value of the $L$ ,
$mvr = \dfrac{{nh}}{{2\pi }}$
Substituting the above equation in the equation (1), we get
$r = \dfrac{{4{\pi ^2}K{e^2}m}}{{{n^2}{h^2}}}$
From the above equation, it is clear that $r\alpha \dfrac{1}{{{n^2}}}$ -------(2)
The total energy is obtained by subtracting the kinetic energy by the potential energy,
$E = \dfrac{1}{2}m{v^2} - \dfrac{{K{e^2}}}{{3{r^2}}} = \dfrac{{K{e^2}}}{{6{r^3}}}$
Hence $E\alpha {n^6}$

Thus the option (A) is correct.

Note:The energy of the atom that is located in the ${n^{th}}$ orbit is calculated by finding the energy required by the electron to move to the other orbit than the energy with it. Since the electron is revolving around the orbit, the angular momentum is used.