Question

In the Bohr's model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If \[{a_0}\]​ is the radius of the ground state orbit, m is the mass and e is charge on the electron and \[{\varepsilon _0}\] ​ is the vacuum permittivity, the speed of the electron is
A. Zero
B. \[\dfrac{e}{{\sqrt {{\varepsilon _0}{a_0}m} }} \\ \]
C. \[\dfrac{e}{{\sqrt {4\pi {\varepsilon _0}{a_0}m} }} \\ \]
D. \[\sqrt {\dfrac{{4\pi {\varepsilon _0}{a_0}m}}{e}} \]

Answer 1

Hint:In the nucleus of Hydrogen atoms protons and neutrons reside. The electron moves in a circular orbit around the nucleus. As neutrons are neutral units, the centripetal force is furnished by the coulomb attraction between the proton and the electron.

Formula used:
\[{F_i} = \dfrac{{{q_1} \times {q_2}}}{{4\pi {\varepsilon _0}{d^2}}}\]
where \[{F_i}\] is the inward coulomb’s force of attraction between charges \[{q_1}\] and \[{q_2}\] separated by distance d.
\[{F_o} = \dfrac{{m{v^2}}}{r}\]
where \[{F_o}\] is the outward centrifugal force acting on mass m revolving around an orbit of radius r with speed v.

Complete step by step solution:
The mass of the electron is given as m. The charge on the electron is e. As we know that the magnitude of charge on protons is the same as the magnitude of charge on electrons, so the charge on protons is also e. The electron is orbiting in the circular orbit of radius \[{a_0}\].

The centripetal force acting on the electron is due to the Coulomb force of attraction between two charges kept at distance equal to the radius of the orbit.
\[{F_i} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}a_0^2}}\]
Let the speed of the electron is v in the circular orbit, then radially outward there will be centrifugal force acting on it.
\[{F_0} = \dfrac{{m{v^2}}}{{{a_0}}}\]
When the electron is performing uniform circular motion in a constant radius of orbit, it is at equilibrium along the radial direction.

So, the outward centrifugal force is balanced by inward centripetal force.
\[{F_i} = {F_o}\]
\[\Rightarrow \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}a_0^2}} = \dfrac{{m{v^2}}}{{{a_0}}} \\ \]
\[\Rightarrow {v^2} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}m{a_0}}} \\ \]
\[\Rightarrow v = \sqrt {\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}m{a_0}}}} \\ \]
\[\therefore v = \dfrac{e}{{\sqrt {4\pi {\varepsilon _0}m{a_0}} }} \\ \]
So, the speed of the electron in the circular orbit about the nucleus of the hydrogen atom is equal to \[\dfrac{e}{{\sqrt {4\pi {\varepsilon _0}{a_0}m} }}\].

Therefore, the correct option is C.

Note: The nucleus of the hydrogen has mass and the electron too has mass. Due to mass there is gravitational force of attraction between the nucleus and the electron which contribute to the centripetal force. But, the magnitude of the gravitational force of attraction between the nucleus of hydrogen and the orbiting electron is insignificant relative to the coulomb’s force of attraction. So, we don’t neglect it while solving the problem.