Question

In a hypothetical system a particle of mass m and charge -3q is moving around a very heavy particle having charge q. Assuming Bohr’s model to be true to this system the orbital velocity of mass m when it is nearest to heavy particle is:
$\begin{align}
  & A.\text{ }\dfrac{3{{q}^{2}}}{2{{\varepsilon }_{0}}h} \\
 & B.\text{ }\dfrac{3{{q}^{2}}}{4{{\varepsilon }_{0}}h} \\
 & C.\text{ }\dfrac{3q}{2{{\varepsilon }_{0}}h} \\
 & D.\text{ }\dfrac{3q}{4{{\varepsilon }_{0}}h} \\
\end{align}$

Answer 1

Hint: When an electron is moving around the nucleus it is experiencing centripetal force which is equal to force between two charges and by equalizing this centripetal force with the equation of angular momentum we will lead to the correct answer.

Formula used:
1) ${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
2) $\Rightarrow F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}_{2}}}$
3) $mvr=\dfrac{nh}{2\pi }$

Complete step by step solution:
$\to $ When a particle is moving in the circular path it is experiencing centripetal force which is equal to attractive force between charges -3q and q
${{F}_{c}}=F....\left( 1 \right)$

$\to $ Where centripetal force $\left( {{F}_{c}} \right)$ is
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}....\left( 2 \right)$
Where, ${{F}_{c}}$ = centripetal force
m = mass, v = velocity, r = radius

And according to the coulomb’s law force between two charges is
 $\begin{align}
  & \Rightarrow F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}_{2}}} \\
 & \therefore F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{3{{q}^{2}}}{{{r}^{2}}}......\left( 3 \right) \\
\end{align}$
F = attractive or repulsive electric force between two point charges
r = distance between charges
k = coulomb’s constant which is equal to $\dfrac{1}{4\pi {{\varepsilon }_{0}}}$

Now substitute value of equation (2) and equation (3) in equation (1)
$\dfrac{m{{v}^{2}}}{r}=\dfrac{3q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$

Now as we have to compare above equation with the angular momentum we will convert as equation of angular momentum
$\begin{align}
  & \Rightarrow \dfrac{mv{{r}^{2}}}{r}=\dfrac{3{{q}^{2}}}{4\pi {{\varepsilon }_{0}}v} \\
 & \therefore mvr=\dfrac{3{{q}^{2}}}{4\pi {{\varepsilon }_{0}}v}....\left( 4 \right) \\
\end{align}$

Now according to Bohr’s atomic model angular momentum is given by
$mvr=\dfrac{nh}{2\pi }$
r = Radius of the ${{n}^{th}}$ orbit
m = Mass of the electron
n = Orbit of the electron
h = Planck's constant

Now n is given as 1 because the value of n in first orbit is 1
$mvr=\dfrac{h}{2\pi }....(5)$

Now equalizing equation (4) and (5) we get
$\begin{align}
  & \Rightarrow \dfrac{3{{q}^{2}}}{4\pi {{\varepsilon }_{0}}v}=\dfrac{h}{2\pi } \\
 & \Rightarrow v=\dfrac{3{{q}^{2}}\times 2\pi }{4\pi {{\varepsilon }_{0}}h} \\
 & \therefore v=\dfrac{3{{q}^{2}}}{2{{\varepsilon }_{0}}h} \\
\end{align}$

Hence the correct option is (A) $\dfrac{3{{q}^{2}}}{2{{\varepsilon }_{0}}h}$ is correct.

Additional information:
The electrostatic force is an attractive and repulsive force between particles due to their electric charges.

Note:
To solve this type of question we have to consider a particle or an object with mass m as electron and then apply Bohr’s atomic model to get the correct answer.