Basic Modern Physics
12.0 Basic Definitions
12.0 Basic Definitions
Ionization Energy(IE):
Minimum energy required to move an electron from ground state to $n = \infty $ is called ionization energy of the atom or ion.
$$\begin{equation} \begin{aligned} {E_{ionization}} = {E_\infty } - {E_n} = - {E_n} = \frac{{13.6{Z^2}}}{{{n^2}}}\ eV \\ \\\end{aligned} \end{equation} $$
Ionization Potential(IP):
Potential difference through which an electron must be accelerated from rest such that it gains a kinetic energy equal to the ionization energy of the atom is called ionization potential of the atom.
$${V_{ionization}} = \frac{{{E_n}}}{e} = \frac{{13.6{Z^2}}}{{{n^2}}}\ V$$
Excitation:
The process of absorption of energy by an electron to jump from a lower energy level to a higher energy level is known as excitation.
Excited State:
States other than ground state are known as excited states.
Example. $n=2$ is 1st excited state, $n=3$ is 2nd excited state etc.
Excitation energy:
It is the energy required to move from ground state to any other excited state.
$${E_{ionization}} = {E_{higher}} - {E_{ground}}$$
Example. First excitation energy of $H =$ ${E_2}$-${E_1}$ $= -3.4-(-13.6) = 10.2$$eV$
Excitation Potential:
Potential difference through which an electron must be accelerated from rest such that it gains a kinetic energy equal to the excitation energy of any state of the atom is called excitation potential of the atom.
Example. First excitation potential of $H= 10.2$ $V$.
$${V_{ionization}} = \frac{{{E_{excitation}}}}{e}$$
Binding Energy:
It is the amount of energy liberated when different constituents of a system are brought infinity to form a system. It is numerically equal to the ionization energy of the atom.
Example. Binding energy of ground state of a He atom is $54.4$ $eV$.
Shortcomings:
- It is only applicable to hydrogen like atoms.
- It couldn't explain why electrons revolving in stationary orbits disobey the law of electrodynamics.
- It couldn't explain the hyperfine structure of spectral lines.More fine deviations in wavelength of a spectral line were observed as technology of spectroscopy advanced.A spectral line was found to consist of very closely spaced lines, which is known as hyperfine structure of spectral line.
- The model couldn't account for the splitting of spectral lines in electric and magnetic fields.
- The model also violates the uncertainty principle in that it considers electrons to have known orbits and locations, two things which can not be measured simultaneously.
Question 14. The energy levels of a hypothetical one electron atom are given by $$\begin{equation} \begin{aligned} {E_n} = - \frac{{18.6}}{{{n^2}}}\ eV,\quad \quad \quad \quad \quad {\text{where}}\quad n = 1,2,3,... \\ \\\end{aligned} \end{equation} $$
(a) Compute the four lowest energy levels an construct the energy level diagram.
(b) What is excitation potential of the state $n=2$?
(c) What wavelength ($\mathop {{\text{ A}}}\limits^0 $) can be emitted when these atoms in the ground state are bombarded by electrons that have been accelerated through a potential difference of 16.2 $V$?
(d) If these atoms are in the ground state, can they absorb radiation having a wavelength of $2000\mathop {{\text{ A}}}\limits^0 $?
(e) What is the photoelectric threshold wavelength of this atom?
Solution: (a) $$\begin{equation} \begin{aligned} {E_1} = - \frac{{18}}{{{1^2}}} = - 18.0\ eV\quad \quad {E_2} = - \frac{{18}}{{{2^2}}} = - 4.5\ eV\quad \quad {E_3} = - \frac{{18}}{{{3^2}}} = - 2.0\ eV\quad \quad {E_4} = - \frac{{18}}{{{4^2}}} = - 1.125\ eV \\ \\\end{aligned} \end{equation} $$
The energy level diagram is shown in $Fig 17$.
(b) The excitation potential of state $n=2$ is: $18.0-4.5=13.5$ $V$.
(c) Energy of the electron accelerated by a potential difference of 16.2 $V$ is 16.2 $eV$. With this energy, the electron can excite the atom from $n=1$ to $n=3$ as
$$\begin{equation} \begin{aligned} \quad \quad {E_4} - {E_1} = 1.125 - \left( { - 18.0} \right) = 16.875\ eV > {\text{ }}16.2\ eV \\ {\text{and }}{E_3} - {E_1} = 2.0 - \left( { - 18.0} \right) = 16.0\ eV < {\text{ }}16.2\ eV \\ \quad \quad {\lambda _{32}} = \frac{{12375}}{{{E_3} - {E_2}}} = \frac{{12375}}{{ - 2.0 - \left( { - 4.5} \right)}} = 4950\mathop {{\text{ A}}}\limits^0 \\ \quad \quad {\lambda _{31}} = \frac{{12375}}{{{E_3} - {E_2}}} = \frac{{12375}}{{16}} = 773\mathop {{\text{ A}}}\limits^0 \\ {\text{and }}{\lambda _{32}} = \frac{{12375}}{{{E_2} - {E_1}}} = \frac{{12375}}{{ - 4.5 - \left( { - 18.0} \right)}} = 917\mathop {{\text{ A}}}\limits^0 \\\end{aligned} \end{equation} $$
(d) No. The energy corresponding to $\lambda = 2000\mathop {{\text{ A}}}\limits^0 $;
$$\begin{equation} \begin{aligned} E = \frac{{12375}}{{2000}} = 6.187\ eV \\ \\\end{aligned} \end{equation} $$
The minimum excitation energy is $13.5 eV$ ($n=1$ to $n=2$).
(e) Threshold wavelength for photo emission to take place from such an atom is,
$$\begin{equation} \begin{aligned} {\lambda _{\max }} = \frac{{12375}}{{18}} = 687.5\mathop {{\text{ A}}}\limits^0 \\ \\\end{aligned} \end{equation} $$
