Question
Using Bohr's total postulates, derive the expression for the total energy of the electron in the stationary states of hydrogen atom.
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Answer
According to Bohr's postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge +e. For an electron moving with a uniform speed in a circular orbit of a given radius, the centripetal force is provided by Coulomb force of attraction between the electron and the nucleus. The gravitational attraction may be neglected as the mass of electron and proton is very small. So,
$m v^2 / r = ke ^2 / r ^2\left(\right.$ where, $\left.k =1 / 4 \pi \varepsilon_0\right)$
or $m v^2= ke ^2 / r$.
where, m = mass of electron, r = radius of electronic orbit, v = velocity of electron
Again, by Bohr's second postulates
$mvr = nh / 2 \pi$
where, $n=1,2,3, \ldots$ or $v=n h / 2 \pi m r$
Putting the value of $v$ in Eq. (i)
$m\left(\frac{n h}{2 \pi m r}\right)^2=\frac{k r^2}{r} \Rightarrow r=\frac{n^2 h^2}{4 \pi^2 k m e^2}$
Kinetic energy of electron ,
$E_K=\frac{1}{2} m v^2=\frac{k e^2}{2 r}\left(\because \frac{m v^2}{r}=\frac{k e^2}{r^2}\right)$
Using Eq(ii), we get
$E_K=\frac{k \kappa^2}{2} \frac{4 \pi^2 k m c^2}{n^2 h^2}=\frac{2 \pi^2 k^2 m c^4}{n^2 h^2}$
Potential energy of electron,
$E_P=-\frac{k(e) \times(e)}{r}=-\frac{k x^2}{r}$
Using Eq(ii), we get
$E_P=-k e^2 \times \frac{4 \pi^2 k m c^2}{n^2 h^2}=-\frac{4 \pi^2 k^2 m e^4}{n^2 h^2}$
Hence, total energy of the electron in the $n ^{ th }$ orbit
$\begin{array}{l}E=E_P+E_K=-\frac{4 \pi^2 k^2 m e^4}{n^2 h^2}+\frac{2 \pi^2 k^2 m e^4}{n^2 h^2} \\
=-\frac{2 \pi^2 k^2 m e^4}{n^2 h^2}\end{array}$
$m v^2 / r = ke ^2 / r ^2\left(\right.$ where, $\left.k =1 / 4 \pi \varepsilon_0\right)$
or $m v^2= ke ^2 / r$.
where, m = mass of electron, r = radius of electronic orbit, v = velocity of electron
Again, by Bohr's second postulates
$mvr = nh / 2 \pi$
where, $n=1,2,3, \ldots$ or $v=n h / 2 \pi m r$
Putting the value of $v$ in Eq. (i)
$m\left(\frac{n h}{2 \pi m r}\right)^2=\frac{k r^2}{r} \Rightarrow r=\frac{n^2 h^2}{4 \pi^2 k m e^2}$
Kinetic energy of electron ,
$E_K=\frac{1}{2} m v^2=\frac{k e^2}{2 r}\left(\because \frac{m v^2}{r}=\frac{k e^2}{r^2}\right)$
Using Eq(ii), we get
$E_K=\frac{k \kappa^2}{2} \frac{4 \pi^2 k m c^2}{n^2 h^2}=\frac{2 \pi^2 k^2 m c^4}{n^2 h^2}$
Potential energy of electron,
$E_P=-\frac{k(e) \times(e)}{r}=-\frac{k x^2}{r}$
Using Eq(ii), we get
$E_P=-k e^2 \times \frac{4 \pi^2 k m c^2}{n^2 h^2}=-\frac{4 \pi^2 k^2 m e^4}{n^2 h^2}$
Hence, total energy of the electron in the $n ^{ th }$ orbit
$\begin{array}{l}E=E_P+E_K=-\frac{4 \pi^2 k^2 m e^4}{n^2 h^2}+\frac{2 \pi^2 k^2 m e^4}{n^2 h^2} \\
=-\frac{2 \pi^2 k^2 m e^4}{n^2 h^2}\end{array}$
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