Question

Using Bohr’s postulates of the atomic model, derive the expression for radius of n^th electron orbit. Hence obtain the expression for Bohr’s radius.

Answer 1

Hint:In order to solve this question recall Bohr’s second postulate of quantization which states that, an electron revolves around the nucleus in orbits and the angular momentum of revolution is an integral multiple of $\dfrac{h}{{2p}}$, where $h$ is the planck’s constant.

Formula used:
According to Bohr’s second postulate of quantization,
$m{v_n}{r_n} = n\dfrac{h}{{2\pi }}$
Where, $m$ is the mass of the electron revolving around the nucleus
${v_n}$ is the speed of electron in the n^th orbit
${r_n}$ is the radius of n^th orbit
$n$ is the principal quantum number ($n = 1,2,3....$)
$h$ is the planck's constant

Complete step by step solution:
According to Bohr’s second postulate of quantization,
$m{v_n}{r_n} = n\dfrac{h}{{2\pi }}$
On squaring both sides, we have
$ \Rightarrow {m^2}{v_n}^2{r_n}^2 = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}}}$ ……………….(i)
Now we know that the Centripetal force is provided by the electrostatic force of attraction then the equation is given by,
$\dfrac{{m{v^2}}}{r} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r^2}}}$ ……………..(ii)
From equation (i) put value of ${v_n}$ in equation (ii), we have
$ \Rightarrow \dfrac{{m{n^2}{h^2}}}{{4{\pi ^2}{m^2}{r_n}^3}} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r_n}^2}}$
On further solving we have
\[ \Rightarrow {r_n} = \dfrac{{{n^2}{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\]
This is the expression for radius of n^th electron orbit.
Now we know that the Bohr’s radius is the radius of the innermost orbit.
Hence, by putting $n = 1$ in the above equation we get the expression for Bohr’s radius.
${r_1} = \dfrac{{{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}$
This is the expression for Bohr’s radius

Note:According to the Bohr’s third postulate, an electron can transition from a non-radiating orbit to another of a lower energy level. In doing so, a photon is emitted whose energy is equal to the energy difference between the two states.
When the electron is revolving in an orbit closest to the nucleus, the energy of the atom is the least or has the largest negative value.