band gap q&a

8/10/2019 band gap Q&A

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Physics 271 - Homework Set 10 - due Wednesday 27 April 2011

From the textbook: Chapter 9: 27 a ; Chapter 11: 8 b, 9.

Hints:(a) The molar mass is the total mass of Avogadros Number (6 .022 1023 ) of atoms.(b) Look at the electronic conguration of the various elements in Table 8.2

9-27: The density of pure copper is 8 .92 103 kg/m 3 , and its molar mass is 63.546 g. Using theexperimental value of the conduction electron density, 8 .47 1028 m 3 , compute the numberof conduction electrons per atom.

Solution: Compute the number density of atoms :

8.92 103 kg/ m3 1mole

0.063546 kg6.022 1023 atoms

mole = 8.45 1028 m 3

Compared to the given number density of conduction electrons , this is very nearly the same.So (within rounding errors) we can conclude that each atom contributes one electron to theconduction band.

11-8: What kind (p-type or n-type) of semiconductor is made if pure germanium is doped with asmall amount of (a) aluminum; (b) selenium ?

Solution: Using Figure 8.2 (p.273), we see that germanium (Ge) has two electrons in its(unlled) 4 p orbital. Considering what happens when you replace a Ge atom:

(a) Al has only one electron in its (unlled) 3 p orbital, so this becomes a p-type semiconductor.

(b) Se has four electrons in its (unlled) 4 p orbital, so this becomes an n-type semiconductor.

11-9 Find the wavelength of the photon needed to cause an electron to jump from the valence tothe conduction band in (a) germanium, with a band gap of 0.67 eV; (b) silicon, with a bandgap of 1.1 eV.

Solution:

The photon must have an energy equal to or greater than the band gap to give to the electron,

E g = hc

(a) = hc/E g = 1.240 103 eV nm/ 0.67 eV = 1850 nm

(b) = hc/E g = 1.240 103 eV nm/ 1.1 eV = 1130 nm


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Additional problems:

1. The following diagrams show the conduction (C) and valence (V) energy bands of threedifferent crystalline substances. Also shown is the level of the Fermi energy ( E F ). Label eachof the three diagrams as being most likely a conductor, an insulator, or a semiconductor.

CONDUCTOR INSULATOR SEMICONDUCTOR

2. Sodium in a metallic form is monovalent (i.e., it contributes one electron per atom to theconduction band). Assume it has a bulk density of 0 .971g/cm 3 and an atomic mass (seeAppendix 8) of 23.0u. Calculate (a) the density of charge carriers and (b) the Fermi energy

for sodium. (Answers: (a) 2.54 1028

electrons/m 3

; (b) 3.15 eV. )Solution:(a) The atomic density is

na = N AM

= (0.971 g/ cm3 )(6.02 1023 atoms / mole)

(23.0 g/ mole)

= 2 .54 1022 atoms / cm3 = 2 .54 1028 electrons / m3

since there is one electron per atom in the free electron gas.

(b)E F =

h2

8me3ne

2 / 3

= (6.626 10 34 J sec)2

8(9 10 31 kg)3(2.54 1028 electrons / m3 )

2 / 3

= 5 .04 10 19 J = 3 .15 eV


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3. From the optical absorption spectrum of a certain semiconductor, one nds that the longestwavelength of radiation absorbed is 1.85 m. What is the energy band-gap of this semicon-ductor? ( Answer: 0.670 eV )

Solution:E g = hf min =

hcmax

= 1.240 103 eV nm

1850 nm = 0.670 eV

(Compare to Problem 11-9 above.)

band gap q&a

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