CBSE 2014, Set 3

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.

Edvaya · Accepted Answer

The correct answer is Subjective. **SOLUTION** According to the postulates of Bohr’s atomic model, electrons can revolve around the nucleus only in those circular orbits for which the angular momentum of the electron is integral multiple of $\frac{h}{2\pi}$, where $h$ is Planck’s constant. Let, $m$ - mass of the electron $v_n$ - velocity of the electron in $n^{th}$ orbit $r_n$ - radius of the $n^{th}$ orbit Then $$mv_n r_n = n\frac{h}{2\pi}$$ $$\Rightarrow mv_n = n\frac{h}{2\pi r_n} \dots (i)$$ where $n$ is called the **principal quantum number** We know that for an electron revolving in a circular orbit, the centripetal force is equal to the electrostatic force of attraction between the electron and the nucleus. $$\frac{mv_n^2}{r_n} = \frac{kZe^2}{r_n^2}$$ $$mv_n^2 = \frac{kZe^2}{r_n} \dots (ii)$$ Multiplying equation (ii) by $m$ and squaring equation (i), we get left hand side equal to $m^2v_n^2$. Hence equating the R.H.S: $$\frac{kmZe^2}{r_n} = \frac{n^2h^2}{4\pi^2 r_n^2}$$ $$	herefore r_n = \frac{n^2h^2}{4\pi^2 kmZe^2}$$ This is the radius of the $n^{th}$ orbit of an Hydrogen like atom. Substituting the value of $k = \frac{1}{4\pi\varepsilon_0}$, we have $$r_n = \left(\frac{n^2}{m}ight)\left(\frac{h}{2\pi}ight)^2 \frac{4\pi\varepsilon_0}{Ze^2}$$ The radius of first orbit ($n = 1$) of Hydrogen, ($Z = 1$) will be $$r_1 = \frac{h^2\varepsilon_0}{\pi me^2} = a_0$$ This is called the Bohr’s radius, represented by $a_0$. Its numerical value is $5.29 	imes 10^{-11} 	ext{ m}$.

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