Relation between velocity of a electron (In the Bohr model of the atom) and the radius So we can derive this expression by equating force of attraction on the electron by the nucleus to the centripetal force acting on the electron, i.e: $$ \frac{KZe^2}{r^2}= \frac{mv^2}{r},$$ where $m$ is the mass and $e$ is the charge on the electron, $Z$ is atomic no. of the $H$-like particle, $K$ is the Coulomb constant  and $v$ is the tangential velocity.
Using this we can conclude that: $$ v^2=\frac{K Ze^2}{mr}$$ which shows that $$v \varpropto \frac{1}{\sqrt{r}}.$$
But also the Bohr's postulate of angular momentum states that :
$$mvr=\frac{nh}{2 \pi}$$
Which makes it $$v \varpropto \frac{1}{r}.$$
Out of the two relations which is the   correct relation ?
 A: A proportionality is meaningful only in a certain context. In the context of a single atom, the quantities $Z,m,e$ are constant and $n,v,r$ are interdependent on each other.
In the context of a single atom, the second proportionality is not correct because the value of $n$ is not constant (different values of $n$ correspond to different orbits).
$$v=\frac{nh}{2πmr}⇒v∝\frac{n}{r}$$
The first proportionality is correct in the context of a single atom of one element; however, if you are analyzing many elements, $Z$ is also a variable. It too becomes incorrect.
A: The first relationship (the attraction force = centripetal force) is "only" correct for circular classical orbits. It does not contain the Plank constant.
In Quantum Mechanical picture the velocity and the radius are not certain, but "fluctuating" around some average (mean) values. Thus there is no relationship between $v$ and $r$. One can obtain a relationship between $v_0$ and $a_0$ instead (or between $v_n$ and $a_n$, if you like).
