Question

The condition of mechanical stability of the electron of Bohr model

Answer 1

Eq.1
\[\frac{ 1 }{ 4\pi \epsilon _{o} } \frac{ Ze ^{2} }{ r ^{2} } = m \frac{ v ^{2} }{ r }\]
where v is the speed of the electron in its orbit, and r is the radius of the orbit. The
left side of this equation is the Coulomb force acting on the electron, and the right side
is ma, where a is the centripetal acceleration keeping the electron in its circular orbit.
Now, the orbital angular momentum of the electron, L = mvr, must be a constant,
because the force acting on the electron is entirely in the radial direction. Applying
the quantization condition,
\[n \frac{ h }{ 2\pi } \]
to
\[L\]
we have
\[mvr = n \frac{ h }{ 2\pi }\]
solving for v and substituting into Eq.1, we obtain
\[Ze ^{2} = 4\pi \epsilon _{o} mv ^{2}r = 4\pi \epsilon _{o}mr \left( \frac{ nh }{ 2mr \pi } \right) = 4\pi \epsilon _{o}\frac{ n ^{2} h ^{2}}{ 4\pi ^{2}mr}\]
so
\[r = 4\pi \epsilon _{o}\frac{ n ^{2} h ^{2}}{ 4\pi ^{2}mZe ^{2} }\]
where n =1,2,3,...
and
\[v = \frac{ nh }{ 2\pi mr } =\frac{ 1 }{ 4\pi \epsilon _{o}} \frac{ 2\pi Ze ^{2} }{ nh }\]