91Ó°ÊÓ

According to de Broglie, Light has dual nature as in some situations it behaves like waves and in others like particles. Electrons, which sometimes exhibits particle nature may also behave like wave in some situations.

When a particle acts like a wave, it carries a wavelength which is known as de Broglie wavelength.

The energy to ionized the atom means jumping of electron from ground level to infinity.

$\begin{array}{l}∆E=E_{∞}-E_{GL}\\ =0-\left(-2eV\right)\\ =20eV\end{array}$

Therefore, the amount energy to ionize the electron is 20 eV.

As the atom absorbs 18eV electron in ground state:

$\begin{array}{l}∆\mathrm{E}={\mathrm{E}}_4-{\mathrm{E}}_1\\ =-2\mathrm{eV}-\left(-2\mathrm{eV}\right)\\ =18\mathrm{eV}\end{array}$

Which means that electron will go to fourth energy state.

Therefore, the possible energies of emitted photons are:

$$\begin{gathered} {\mathrm{E}}_4-{\mathrm{E}}_3=-2+5=3\mathrm{eV} \\ {\mathrm{E}}_4-{\mathrm{E}}_2=-2+10=8\mathrm{eV} \\ {\mathrm{E}}_4-{\mathrm{E}}_1=-2+20=18\mathrm{eV} \\ {\mathrm{E}}_3-{\mathrm{E}}_2=-5+10=5\mathrm{eV} \\ {\mathrm{E}}_3-{\mathrm{E}}_1=-5+20=15\mathrm{eV} \\ {\mathrm{E}}_2-{\mathrm{E}}_1=10+20=10\mathrm{eV} \end{gathered}$$

As the next energy level is E = -10 eV the photon can’t be absorbed

$\begin{array}{l}{\mathrm{E}}_{\mathrm{n}}=-10\\ ∆\mathrm{E}=-20+10=10\mathrm{eV}\end{array}$

The photon doesn’t have sufficient energy.

Therefore, the photon will not be absorbed.

The energy difference is given by

$\begin{array}{l}∆\mathrm{E}={\mathrm{E}}_{\mathrm{Photon}}\\ ∆\mathrm{E}={\mathrm{E}}_3-{\mathrm{E}}_2=-5+10=5\mathrm{eV}\\ ∆\mathrm{E}={\mathrm{E}}_3-{\mathrm{E}}_1=-5+20=15\mathrm{eV}\end{array}$

For no ejection of electrons

$\mathrm{Δ·¡}={\mathrm{E}}_4-{\mathrm{E}}_3=-2+5=3\mathrm{eV}$

Therefore, the minimum value of work function is 3 eV and maximum value is 5 eV.