Question

For a hypothetical hydrogen like atom, the potential energy of the system is given by $U(r) = \dfrac{{ - K{e^{\mathbf{2}}}}}{{{r^{\mathbf{3}}}}}$ , where $r$ is the distance between the two particles. If Bohr's model of quantization of angular momentum is applicable, then velocity of particle is given by:
(A) $v = \dfrac{{{n^2}{h^3}}}{{K{e^2}8{\pi ^3}{m^2}}}$
(B) $v = \dfrac{{{n^3}{h^3}}}{{24K{e^2}{\pi ^3}{m^2}}}$
(C) $v = \dfrac{{{n^3}{h^3}}}{{8K{e^2}{\pi ^3}{m^2}}}$
(D) $v = \dfrac{{{n^2}{h^3}}}{{24K{e^2}{\pi ^3}{m^2}}}$

Answer 1

Hint: We need to understand the Bohr’s atomic model and its postulates. Niels Bohr gave a model for an atom in order to explain the line spectrum and the stability of an atom. We have to remember that the dormant energy is an energy which is connected with the attractive force and repulsive force between objects.

Complete step by step answer:
Given in the question:
The potential energy of the system containing hydrogen atom $U(r) = \dfrac{{ - K{e^{\mathbf{2}}}}}{{{r^{\mathbf{3}}}}}$ where $r$ is the distance between the two particles. We are to find the velocity of a particle applying Bohr's model of quantization of angular momentum.
The change in the potential energy with respect to the radius will be the force due to which the electron in a system containing hydrogen will be quantized, i. e F=$\dfrac{{d\left[ {U(r)} \right]}}{{dr}} = \dfrac{{3K{e^2}}}{{{r^4}}}$.
From the first postulate, = $\dfrac{{3K{e^2}}}{{{r^4}}}$=\[\dfrac{{m{v^2}}}{r}\].
From the second postulate, \[mvr\]$ = n\dfrac{h}{{2\pi }}$
\[r\]$ = n\dfrac{h}{{2\pi mv}}$
We substitute the value of \[r\]$ = n\dfrac{h}{{2\pi mv}}$ in the equation $\dfrac{{3K{e^2}}}{{{r^4}}}$=\[\dfrac{{m{v^2}}}{r}\] and find the value of velocity$v$.
Therefore, $v = \dfrac{{{n^3}{h^3}}}{{24K{e^2}{\pi ^3}{m^2}}}$
Hence the correct option is option (B).

Additional information:
His model was based on three postulates discussed below:
The electrons revolve around the nucleus only in certain circular orbits called energy shells or energy levels. An electron revolving in a particular energy shell is associated with a fixed amount of energy. The necessary centripetal force required for the circular motion is provided by the electrostatic force of nucleus and negatively charged electron. To further explain this, hydrogen atom was considered as it consists of a single electron. Considering the electron having mass “\[m\]” revolving with speed “\[v\]” in an orbit of radius “\[r\]”. The charge of the nucleus is positive (\[ + e\]) which makes the atom electrically neutral and hence stable. The force of attraction between electron and nucleus is given by Coulomb’s law as:$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{e^2}}}{{{r^2}}}$ . The centripetal force is given by \[\dfrac{{m{v^2}}}{r}\] . Since the centripetal force is responsible for the electron to be moving in its orbit, therefore centripetal force is equal to the force of attraction, i. e, \[\dfrac{{m{v^2}}}{r} = \dfrac{1}{{4\pi {\varepsilon _0}}} \times \dfrac{{{e^2}}}{{{r^2}}}\].
The second postulate states that the electron revolves around only in those orbits for which the angular momentum is an integral multiple of $\dfrac{h}{{2\pi }}$ where $h$ is the Planck’s constant \[\left( {6.6{\text{ }} \times {\text{ }}{{10}^{ - 34}}Js} \right).\]Thus the angular momentum (L) of the revolving electron is stable and does not radiate energy. Thus, ${L_n} = n\dfrac{h}{{2\pi }}$, where n is the orbit number that is the principal quantum number. Also, angular momentum=Iω, where $I = m{r^2}$is the moment of inertia of the electron and ω is the angular velocity whose value is $\dfrac{v}{r}$.
Therefore, $I = m{r^2} \times \dfrac{v}{r} = mvr$.
Hence, \[mvr = n\dfrac{h}{{2\pi }}\].
An electron can make a transition from a stationary state of higher energy ${E_2}$ to a state of lower energy ${E_1}$ and in doing so, it emits a single photon of frequency. Suppose an electron jumps from \[nth\] orbit of higher energy to $pth$ orbit of higher energy, then the energy radiated will be ${E_{nth}} - {E_{pth}}$.

Note:
Note that Bohr’s postulates are applicable for only one electron system only. The energy associated with a stationary state of an electron only depends on $n$(principal quantum number) for one electron system. For a multi electron system, this is not valid. Also it provides an incorrect value for the ground state orbital angular momentum. The drawbacks of this model should be kept in mind.