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

(a). The Fermi energy is
`5.75*10^(-17)\ eV`

(b). The next available energy level in a 1 cm³ is
`1.51*10^(-6)\ eV`

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

Step-by-step explanation:

Given that,

Number of electron
`n=18*10^(2)\ electron/m^3`

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

Using formula of Fermi energy

`E_(f)=(h)/(8m)((3)/(\pi))^{(2)/(3)}n^{(2)/(3)}`

Put the value into the formula

`E_(f)=((6.67*10^(-34))^2)/(8*9.1*10^(-31))((3)/(\pi))^{(2)/(3)}(18*10^(2))^{(2)/(3)}`

`E_(f)=5.75*10^(-17)\ eV`

The Fermi energy is
`5.75*10^(-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)=(n^2\pi^2\hbar^2)/(2ma^2)`

Put the value into the formula

`E_(n)=((2)^2*\pi^2*(1.0545*10^(-34))^2)/(2*9.1*10^(-31)*(10^(-6))^2)`

`E_(n)=0.0000015075\ ev`

`E_(n)=1.51*10^(-6)\ eV`

The next available energy level in a 1 cm³ is
`1.51*10^(-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)=(n^2\pi^2\hbar^2)/(2ma^2)`

Put the value into the formula

`E_(n)=((2)^2*\pi^2*(1.0545*10^(-34))^2)/(2*9.1*10^(-31)*(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.