Basic Modern Physics
11.0 Energy of electron in the $n^{th}$ orbit
11.0 Energy of electron in the $n^{th}$ orbit
There are two kinds of energy associated with a revolving electron, namely kinetic energy (due to velocity of circular motion) and potential energy (due to electrostatic force of attraction).
Kinetic energy:
$$\begin{equation} \begin{aligned} \frac{{m{v^2}}}{r} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{(Ze)(e)}}{{{r^2}}}\quad \Rightarrow \quad \frac{1}{2}m{v^2} = \frac{{Z{e^2}}}{{8\pi {\varepsilon _0}r}} \\ \quad \quad \quad \quad \quad \quad {\text{or}}\;KE = \frac{{Z{e^2}}}{{8\pi {\varepsilon _0}r}} \\\end{aligned} \end{equation} $$
Potential energy:
$$\begin{equation} \begin{aligned} PE = - \frac{{Z{e^2}}}{{4\pi {\varepsilon _0}r}} \\ \\\end{aligned} \end{equation} $$
Total energy:
$$\begin{equation} \begin{aligned} E = KE + PE = \frac{{Z{e^2}}}{{8\pi {\varepsilon _0}r}} - \frac{{Z{e^2}}}{{4\pi {\varepsilon _0}r}}r \\ {\text{or }}E = - \frac{{Z{e^2}}}{{8\pi {\varepsilon _0}r}} \\\end{aligned} \end{equation} $$
So, $$E = - KE = \frac{1}{2}PE$$
Using the previously derived value of $r$, we have
$$\begin{equation} \begin{aligned} E = - \frac{{Z{e^2}}}{{8\pi {\varepsilon _0}r}} = - \frac{{Z{e^2}}}{{8\pi {\varepsilon _0}\left( {\frac{{{n^2}{h^2}{\varepsilon _0}}}{{\pi m{e^2}Z}}} \right)}} = - \left( {\frac{{m{e^4}}}{{8{h^2}\varepsilon _0^2}}} \right)\frac{{{Z^2}}}{{{n^2}}} = - \left( {13.6} \right)\frac{{{Z^2}}}{{{n^2}}}{\text{ev}} \\ {\text{Also, }}E = - \left( {\frac{{m{e^4}}}{{8{h^2}\varepsilon _0^2}}} \right)\frac{{{Z^2}}}{{{n^2}}} = - \left( {\frac{{m{e^4}}}{{8c{h^3}\varepsilon _0^2}}} \right)ch\frac{{{Z^2}}}{{{n^2}}} = - \left( R \right)\left( {ch} \right)\frac{{{Z^2}}}{{{n^2}}} \\ R = Ryderberg's\ constant = \left( {\frac{{m{e^4}}}{{8c{h^3}\varepsilon _0^2}}} \right) = 1.097 \times {10^7}{m^{ - 1}} \\ Rch = Ryderberg's\ energy = 2.17 \times {10^{ - 18}}J = 13.6\ eV \\\end{aligned} \end{equation} $$
The significance of the negative sign is that the electron is bound to the nucleus by force of attraction and external energy 
must be supplied to separate the electron from the nucleus. Hence, the system is said to be binded.
must be supplied to separate the electron from the nucleus. Hence, the system is said to be binded.Now, the frequency of the emitted radiation due to an electron transition from ${n_2}$ to ${n_1}$ is given as,
$$\begin{equation} \begin{aligned} f = \frac{{{E_2} - {E_1}}}{h} = - {Z^2}Rc\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right) \\ or\ \bar v(wave\ number) = \frac{1}{{\lambda \left( {wavelength} \right)}} = \frac{f}{c} = - {Z^2}R\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right) \\\end{aligned} \end{equation} $$
All the equations derived are valid only for H-like atoms (like ${\text{H}}\left( {{\text{Z = 1}}} \right){\text{, H}}{{\text{e}}^ + }\left( {Z = 2} \right),{\text{ L}}{{\text{i}}^{ + + }}\left( {Z = 3} \right){\text{ etc}}{\text{.}}$).
In brief,
$$\begin{equation} \begin{aligned} {r_n} = (0.53)\frac{{{n^2}}}{Z}\mathop {\text{A}}\limits^0 \\ {v_n} = \left( {2.19 \times {{10}^6}} \right)\frac{Z}{n}\ m{s^{ - 1}} \\ {E_n} = - \left( {13.6} \right)\frac{{{Z^2}}}{{{n^2}}}\ eV \\\end{aligned} \end{equation} $$
Question 11. What is the angular momentum of an electron in Bohr's hydrogen atom whose energy is -3.4 $eV$.
Solution: First, we need to identify the quantum number of the energy level.
$$\begin{equation} \begin{aligned} {\text{As }}E = - \frac{{13.6}}{{{n^2}}},{\text{ we have}} \\ \quad \quad - 3.4 = - \frac{{13.6}}{{{n^2}}}\ \ {\text{or}}\ \ n = 2 \\ {\text{The angular momentum quantization gives, }}L = mvr = \frac{{nh}}{{2\pi }} \\ {\text{Substituting }}n = 2,{\text{ we get }}L = \frac{{2h}}{{2\pi }}{\text{ = }}\frac{h}{\pi } \\\end{aligned} \end{equation} $$
Question 12. Find the quantum number $n$ corresponding to the excited state of $H{e^ + }$ ion if on transition to the ground state that ion emits two photons in succession with wavelength $1026.7\mathop {{\text{ A}}}\limits^0 $ and $304\mathop {{\text{ A}}}\limits^0 $. $\left( {R = 1.096 \times {{10}^7}{\text{ }}{m^{ - 1}}} \right)$
Solution: $$\begin{equation} \begin{aligned} {\text{Given }}\frac{1}{{{\lambda _1}}} + \frac{1}{{{\lambda _2}}} = R{Z^2}\left( {1 - \frac{1}{{{n^2}}}} \right) \\ \quad \frac{1}{{{n^2}}} = 1 - \left[ {\frac{{{\lambda _1} + {\lambda _2}}}{{{\lambda _1}{\lambda _2}}} \times \frac{1}{{R{Z^2}}}} \right] \\ \quad \quad \;\; = 1 - \frac{{1330.7 \times {{10}^{ - 10}}}}{{1026.7 \times 304 \times {{10}^{ - 20}} \times 4 \times 1.096 \times {{10}^7}}} \\ {\text{Thus, }}\frac{1}{{{n^2}}} = 0.0275\quad \Rightarrow \quad n = 6.03 \\ {\text{Hence, the quantum number}} = 6 \\\end{aligned} \end{equation} $$
