Question
Using the formula for the radius of $n ^{\text {th }}$ orbit $r_n=\frac{n^2 h^2 \varepsilon_0}{\pi m e^2}$ derive an expression for the total energy of electron in $n^{\text {th }}$ Bohr's orbit.

Answer

From Bohr's second postulate, the formula for the radius of $n^{\text {th }}$ orbit for hydrogen atom is.
$r_n=\frac{n^2 h^2 \varepsilon_0}{\pi m e^2}$
The total energy of the electron in the stationary states of the hydrogen atom is
$E _n=-\frac{1}{8 \pi \varepsilon_0} \frac{e^2}{r_n}$
Using equation (1) and equation (2)
$E _n=-\frac{1}{8 \pi \varepsilon_0} \frac{e^2}{\left(\frac{ n ^2 h^2 \kappa_0}{\pi m e^2}\right)}$
$\therefore E _n=-\frac{m e^4}{8 \varepsilon_0{ }^2 n^2 h^2}$
Substituting $m=9.1 \times 10^{-31} kg$
$\begin{array}{l}
e=1.6 \times 10^{-19} C \\
\varepsilon_0=8.85 \times 10^{-12} \frac{ C ^2}{ Nm ^2} \\
h=6.625 \times 10^{-34} Js
\end{array}$
Simplifying equation,
$E _n=-\frac{2.18 \times 10^{-18}}{n^2} J$
Atomic energies are often expressed in electron volts $(e V)$.
$\begin{array}{l} 
\therefore E _n=-\frac{2.18 \times 10^{-18}}{n^2 \times 1.6 \times 10^{-19}} eV \\
\therefore E _n=-\frac{13.6}{n^2} eV
\end{array}$
The negative sign of the total energy of an electron moving in an orbit means that the electron is bound with the nucleus.
Thus, energy will be required to remove the electron from the hydrogen atom to a distance infinitely far away from its nucleus.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free