Question

Using postulates of Bohr's theory of hydrogen atom, show that
(i) Radii of orbits increases as $n^2$ and
(ii) The total energy of electron increases as $\frac{1}{n^2}$, where $n$ is the principal quantum number of the atom.

Answer 1

Angular momentum as per Bhor Theory $L=mvr=\frac{nh}{2\pi }$-----(A)

Thus velocity $v=\frac{nh}{2\pi mr}$

Now squaring the above equation we get

$v^2=\frac{n^2{h}^2}{4{\pi }^2{m}^2{r}^2}$-------(1)

Also centripetal force=electrostatic force

$\frac{mv}{r}=\frac{e^2}{4\pi E_0{r}^2}$

$v^2=\frac{r{e}^2}{m4\pi E_0{r}^2}=\frac{e^2}{4\pi E_{0}rm}$-------(2)

Now putting the value of 1 in 2 we get

$\frac{n^2{h}^2}{4{\pi }^2{m}^2{r}^2}=\frac{e^2}{4\pi E_{0}mr}$

$r=\frac{n^2{E}_0{h}^2}{\pi e^{2}m}$-------(3)

From equation 3 we observe that radius increases with $n^2$

Thus putting the value of 3 in A we get,

$v=\frac{\frac{nh}{2\pi }}{m\left(\frac{n^2{E}_0{h}^2}{\pi e^{2}m}\right)}$

$v=\frac{e^2}{2nh{E}_0}$--------(4)

Again $E=m{v}^2$

Replacing the value of v we get,

$E=m{\left(\frac{e^2}{2nh{E}_0}\right)}^2=m\left(\frac{e^2}{4n^2{h}^2{E_0}^2}\right)$

Thus from the above equation we observe that total energy E of an electron increases by $\frac{1}{n^2}$