91Ó°ÊÓ

• Zinc is a bivalent metal, ${\mathrm{n}}_{\mathrm{conduction}}=2{\mathrm{n}}_{\mathrm{atom}}$
• Density of the zinc, $\mathrm{d}=7.133\mathrm{g}/{\mathrm{cm}}^3$
• Molar mass of the zinc, A = 65.37 g/mol

Using the given formula for the number density, we can get the required value of the conduction electrons per unit volume, considering the metal is bivalent. Now using this value in the equation of Fermi energy, we calculated the energy of the metal. Now, using this value of Fermi energy, we can get the speed of the electron in conduction band. Further, we can use this value of speed in the de-Broglie wavelength relation and get the value of the wavelength of the electron.

Formulae:

The number density of atoms per unit volume in a material,

$n=\frac{2d}{A/N_A}............................\left(1\right)$

where${\mathrm{N}}_{\mathrm{A}}=6.022×{10}^{23}/\mathrm{mol}$

The equation of Fermi energy

$E_F=\frac{0.121h^2}{m_e}{n}^{2/3}.....................\left(2\right)$

Where n is the number of conduction electrons per unit volume, $m_e$is the mass of an electron and is the Planck’s constant.

The energy of a system of a moving body,

$E=\frac{1}{2}{m}_e{v}_F^2......................................\left(3\right)$

The de-Broglie wavelength of the moving body,

$\mathrm{λ}=\frac{h}{m_e{v}_F}......................................\left(4\right)$

Here $v_F$is the Fermi speed of electrons.

Using the given data in equation (1), we can get the number density of conduction electrons in zinc (considering that zinc is bivalent) as follows:

$$\begin{gathered} \mathrm{n}=\frac{2×7.133\mathrm{g}/{\mathrm{cm}}^3×\left(6.022×{10}^{23}/\mathrm{mol}\right)}{\left(65.37\mathrm{g}/\mathrm{mol}\right)} \\ =1.31×{10}^{29}{\mathrm{m}}^{-3} \end{gathered}$$
Hence, the value of the number density is $1.31×{10}^{29}{\mathrm{m}}^{-3}$.

Using the above number density value in equation (2), we can get the required value of the Fermi energy as follows:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{F}}=\frac{0.121{\left(6.63×{10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{\left(9.11×{10}^{-31}\mathrm{kg}\right)\left(1.6×{10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{\left(1.31×{10}^{29}{\mathrm{m}}^{-3}\right)}^{2/3} \\ =9.42\mathrm{eV} \end{gathered}$$

Hence, the value of the Fermi energy is 9.42 eV.

We can equation the Fermi energy to energy of the light,$m_e{c}^2$

Using the above energy value in equation (3) and equating this to the above energy value, we can get the value of the Fermi speed of the conduction electrons in zinc as follows:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{F}}=\sqrt{\frac{2{\mathrm{E}}_{\mathrm{F}}{\mathrm{c}}^2}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{c}}^2}} \\ =\sqrt{\frac{2\left(9.42\mathrm{eV}\right){\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}^2}{511×{10}^3\mathrm{eV}}} \\ =1.82×{10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the Fermi speed is $1.82×{10}^6\mathrm{m}/\mathrm{s}$.

Using the above speed value in equation (4), we can get the value of the de-Broglie wavelength as follows:

$$\begin{gathered} {\mathrm{λ}}_{\mathrm{F}}=\frac{\left(6.63×{10}^{-34}\mathrm{J}.\mathrm{s}\right)}{\left(9.11×{10}^{-31}\mathrm{kg}\right)\left(1.82×{10}^6\mathrm{m}/\mathrm{s}\right)} \\ =0.40×{10}^{-9}\mathrm{m} \\ =0.40\mathrm{nm} \end{gathered}$$

Hence, the value of the wavelength is 0.40 nm.