Question

Al has 18 x 102s electrons/m'. (a) Find its Fermi energy in electron volts. b) Find the s 18 x 102 electrons/m3. (a) Find its Fe next available energy level in a 1 cm3 crystal of Al. () Do the same for a 10 nm cube quantum dot.

Answer 1

Answer:

(a). The Fermi energy is $5.75\times10^{-17}\ eV$

(b). The next available energy level in a 1 cm³ is $1.51\times10^{-6}\ eV$

(c). The next available energy level in a 10 nm is 1.508 eV.

Explanation:

Given that,

Number of electron $n=18\times10^{2}\ electron/m^3$

(a). We need to calculate the Fermi energy in electron volts

Using formula of Fermi energy

$E_{f}=\dfrac{h}{8m}(\dfrac{3}{\pi})^{\frac{2}{3}}n^{\frac{2}{3}}$

Put the value into the formula

$E_{f}=\dfrac{(6.67\times10^{-34})^2}{8\times9.1\times10^{-31}}(\dfrac{3}{\pi})^{\frac{2}{3}}(18\times10^{2})^{\frac{2}{3}}$

$E_{f}=5.75\times10^{-17}\ eV$

The Fermi energy is $5.75\times10^{-17}\ eV$

(b). We need to calculate the next available energy level in a 1 cm³

$a = 1 cm^3 = 10^{-6}\ m^3$

n = 2

Using formula of energy

$E_{n}=\dfrac{n^2\pi^2\hbar^2}{2ma^2}$

Put the value into the formula

$E_{n}=\dfrac{(2)^2\times\pi^2\times(1.0545\times10^{-34})^2}{2\times9.1\times10^{-31}\times(10^{-6})^2}$

$E_{n}=0.0000015075\ ev$

$E_{n}=1.51\times10^{-6}\ eV$

The next available energy level in a 1 cm³ is $1.51\times10^{-6}\ eV$.

(c). We need to calculate the next available energy level in a 10 nm

$a = 10 nm = 10^{-9}\ m^3$

n = 2

Using formula of energy

$E_{n}=\dfrac{n^2\pi^2\hbar^2}{2ma^2}$

Put the value into the formula

$E_{n}=\dfrac{(2)^2\times\pi^2\times(1.0545\times10^{-34})^2}{2\times9.1\times10^{-31}\times(10^{-9})^2}$

$E_{n}=1.508\ ev$

The next available energy level in a 10 nm is 1.508 eV.

Hence, This is the required solution.