Question

Using Bohr’s atomic model, derive an equation for the radius of orbit of an electron.

Answer 1

In the Bohr model of the particle, electrons travel in characterized circles around the core. The circles are marked by a number, the quantum number n. Electrons can bounce starting with one circle then onto the next by transmitting or engrossing energy.
We realize that the Bohr Model has a few mistakes, however, utilized on account of its basic way to deal with the nuclear hypothesis.
The Bohr model was additionally the primary nuclear model to fuse the quantum hypothesis, implying that it's the archetype of the present more exact quantum-mechanical models.
The Bohr model turns out just for hydrogen since it considers just the connections between one electron and the core.
The Bohr model depends on the energy levels of one electron circling a core at different energy levels. Some other electrons in the molecule will repulse the one-electron and change its energy level
Formula used:
$l = \dfrac{{nh}}{{2\pi }}$
$l =$angular momentum
$n =$principal quantum number
$h =$planck’s constant

Complete step by step solution:
Let,
$mvr = \dfrac{nh}{2 \pi}$...........(1)
$\dfrac{mv^2}{r}= \dfrac{kZe^2}{r^2}$.........(2)
Cancel out $r$ on both sides
$\Rightarrow r = \dfrac{{Kz{e^2}}}{{m{v^2}}}$from $(2)$
$\Rightarrow V = \dfrac{{nh}}{{2\pi \times mr}}$from $(1)$
Put $'V'$,$'m'$,$'r'$in $(2)$
$\Rightarrow r = \dfrac{{Kz{e^2}}}{m} \times \dfrac{{{{(2\pi )}^2}{m^2}{r^2}}}{{{{(nh)}^2}}}$
Taking reciprocal
$\Rightarrow r = \dfrac{{m{{(nh)}^2}}}{{Kz{e^2}(4{\pi ^2}){m^2}}}$
Cancel out m,
$\Rightarrow r = (\dfrac{{{h^2}}}{{mKz{e^2}4{\pi ^2}}})\dfrac{{{n^2}}}{z}$

Note: These types of questions are directly from the theory of bohr’s model. electrons revolve around the nucleus in a circular path due to centripetal force provided by electrostatic attraction between electrons and protons.