Question

Using Bohr's postulates, derive the expression for the total energy of the electron revolving in nth orbit of hydrogen atom. Find the wavelength of $H_{\alpha }$ line, given the value of Rybderg constant, $R=1.1\times {10}^7{m}^{-1}$

Answer 1

We know,

$mvr=\frac{nh}{2n}$ (i) [Bhor's theorem]

Also,

$\frac{m{v}^2}{r}=\frac{kZ{e}^2}{r^2}$ - (ii)

We know in circular motion about electrostatic field

$PE=-2KE$

$TE=PE+KE=-2KE+KE$

$TE=-KE$

$KE=\frac{m{v}^2}{2}$

From (i) and (ii), we have

$r=\frac{nh}{2nmv}$

So, $m{v}^2=\frac{kz{e}^2}{r}=\frac{kz{e}^2}{\frac{nh}{2nmv}}$

$\Rightarrow m{v}^2=\frac{2nmvkz{e}^2}{nh}$

$\Rightarrow v=\frac{2nkz{e}^2}{nh}$

So, $\frac{m{v}^2}{2}=\frac{m}{2}{\left[\frac{2nkz{e}^2}{nh}\right]}^2$

$\frac{m{v}^2}{2}=\frac{2n^2{k}^2{z}^2{e}^{2}m}{n^2{h}^2}$

$\frac{m{v}^2}{2}=\left[\frac{2n^2{k}^2{e}^{4}m}{h^2}\right]\left(\frac{z^2}{n^2}\right)$

So, $TE=-KE=\frac{-m{v}^2}{2}=\left[\frac{-2n^2{k}^2{e}^{4}m}{h^2}\right]\left(\frac{z^2}{n^2}\right)$

We know,

For $H\alpha$ line,

$n_1=2$ and $n_2=3$

$\frac{1}{\lambda }=R\left(z^2\right)[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{1}{\lambda }=(1.1\times {10}^7)\left(1\right)[\frac{1}{4}-\frac{1}{9}]$

$\frac{1}{\lambda }=0.513\times {10}^7=1.53\times {10}^8{m}^{-1}$

$\lambda =\frac{1}{153\times {10}^8}=65.35A$