Question

Find the radius of ${{2}^{\text{nd}}}$ and ${{3}^{\text{rd}}}$ Bohr orbit of the hydrogen atom. ($\text{m=9}\text{.1x1}{{\text{0}}^{31}}\text{Kg}$,$\text{e=1}\text{.6x1}{{\text{0}}^{19}}\text{C}$)

Answer 1

Hint: Radius of Bohr’s orbit can be calculated by the formula. It is proportional to n th orbit and inversely proportional to square of atomic number. This formula works for single electron species.

Complete step by step answer:
> Radius of Bohr’s orbit in hydrogen and hydrogen like species can be calculated by using the following formula.
- Radius of Orbit, \[\text{r=}\dfrac{{{\text{n}}^{2}}{{\text{h}}^{2}}}{4{{\pi }^{2}}m{{e}^{2}}}\text{x}\dfrac{1}{Z}=0.529\text{x}\dfrac{{{\text{n}}^{2}}}{Z}\]
where, n= principal quantum number of orbit.
Z= atomic number
m=Mass of electron=$9.1\text{x1}{{\text{0}}^{-31}}\text{kg}$
e= charge of electron=$\text{1}\text{.6x1}{{\text{0}}^{-19}}\text{C}$
n= number of orbit
h= Planck’s constant
> Bohr proposed that the allowed orbits are circular and must have quantized orbital angular momentum given by-
$\text{L=mvr=n}\dfrac{\text{h}}{2\pi }$$(\text{n=1,2,3}...\text{)}$
- For 2nd radius of the hydrogen atom,
\[{{\text{r}}_{2}}=0.529\dfrac{{{\text{n}}^{2}}}{{{Z}^{2}}}=0.529\dfrac{{{(2)}^{2}}}{{{1}^{2}}}=0.529\text{x4=2}\text{.116}\]
- For 3rd radius of the hydrogen atom,
\[{{\text{r}}_{3}}=0.529\dfrac{{{\text{n}}^{2}}}{{{Z}^{2}}}=0.529\dfrac{{{(3)}^{2}}}{{{1}^{2}}}=0.529\text{x9=4}\text{.761}\]
Furthermore, this formula is valid for hydrogen-like species i.e. it should have one electron in its shell. The Bohr Theory gives accurate values for the energy levels in hydrogen-like atoms, but it has been improved upon in several respects.
Hence, the radius of 2nd and 3rd Bohr orbit in a hydrogen atom is 2.116 and 4.761 Angstrom.

Note: There are certain limitations to Bohr’s theory. It cannot be applied to multi-electron atoms, even one as simple as a two-electron helium atom. Bohr’s model is what we call semi-classical. The orbits are quantized (non-classical) but are assumed to be simple circular paths (classical). As quantum mechanics was developed, it became clear that there are no well-defined orbits; rather, there clouds of probability.