Question

Obtain An Expression For The Radius Of Bohr Orbit.

Answer 1

Step 1: Calculation of Speed of moving electron

1. According to Bohr second postulate the radius can only take values such that the angular momentum of the electron in these orbits is an integral multiple of $\frac{\mathrm{nh}}{2\mathrm{\pi }}$.
2. From Bohr's second postulate ${\mathrm{L}}_{\mathrm{n}}={\mathrm{mv}}_{\mathrm{n}}{\mathrm{r}}_{\mathrm{n}}=\frac{\mathrm{nh}}{2\mathrm{\pi }}$ where n is an integer, ${\mathrm{r}}_{\mathrm{n}}$ is the radius of n^th orbit, v_n is the speed of moving electron in the n^th orbit.
3. So the velocity of moving electron ${\mathrm{v}}_{\mathrm{n}}=\frac{\mathrm{nh}}{2{\mathrm{\pi mr}}_{\mathrm{n}}}$.

Step 2: Calculation of radius of Bohr's orbit

1. The relation between velocity of moving electron and radius will be: ${\mathrm{v}}_{\mathrm{n}}=\frac{\mathrm{e}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{mr}}_{\mathrm{n}}}$
2. Substituting the value of v_n from the above we get: $\frac{\mathrm{nh}}{2{\mathrm{\pi mr}}_{\mathrm{n}}}=\frac{e}{4{\mathrm{\pi \epsilon }}_0{\mathrm{mr}}_{\mathrm{n}}}$
3. After solving for r we get: $\mathrm{r}=\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\mathrm{\pi }}^2{\mathrm{me}}^2}\times \frac{1}{\mathrm{z}}$ or $\mathrm{r}=0.529\times \frac{{\mathrm{n}}^2}{\mathrm{z}}\stackrel{°}{\mathrm{A}}$

Therefore, the radius of Bohr's orbit is $\mathrm{r}=0.529\times \frac{{\mathrm{n}}^2}{\mathrm{z}}\stackrel{°}{\mathrm{A}}$.