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unscientific

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## Homework Statement

[/B]

(a) Explain the terms intrinsic, extrinsic, mobility and effective mass in semiconductors

(b) What is a hole and explain its mass and charge

(c) Why is mobility of holes often less than mobility of electrons? Find the number density of holes and electrons

(d) Find the mobility of the metal

(e) Find the drift velocities of electrons in the metal and germanium

Only major problem I have is part (e).

## Homework Equations

## The Attempt at a Solution

__Part(a)__[/B]

Intrinsic: No impurities, density of holes = density of electrons

Extrinsic: Impurities present, N-doping or P-doping

Mobility: Ease of movement of electrons and holes through semiconductor ##\mu = \frac{v}{E}##.

Effective mass: mass near the bottom of Conduction band or top of valence band

__Part(b)__A hole is an absence of an electron. It has opposite charge, and opposite velocity to the electron, since overall charge must be conserved. It has the same effective mass of electrons, due to conservation of momentum.

__Part(c)__Holes are surrounded by a sea of bounded electrons in the valence band whereas free electrons are not as exposed in the conduction band due to the absence of states. So it is easier for free electrons to move around inhibited compared to holes.

Using ##J = nev = \sigma E##, we have ##\frac{v}{E} = \frac{\sigma}{ne} = \frac{1}{ne\rho}##.

[tex]n_e = \frac{1}{e \rho \mu_e} = 6.94 \times 10^{19} m^{-3} [/tex]

[tex]n_h = \frac{1}{e \rho \mu_h} = 3.47 \times 10^{19} m^{-3} [/tex]

__Part(d)__The number density for this metal is ##n = \frac{N}{a^3} = \frac{4}{a^3} = 8.57 \times 10^{28} m^{-3}##.

[tex]\mu_{metal} = 4.28 \times 10^{-3}[/tex]

It seems that the mobility in this metal is about 100 times less than the mobility in germanium.

__Part(e)__Since ##v_d = \mu E##, for a given mobility the drift velocity is dependent on electric field. But since the electric field is ##E = \frac{V}{l}##, if we do not know the length of the metal, how could we figure out the voltage or eletric field across it? Surely an infinitely long metal would dominate the voltage across it than the germanium.

[tex]v = \mu E = \mu \left(\frac{V}{l}\right)\left( \frac{\rho l}{A} \right) = \frac{\mu V \rho}{A} [/tex]

Can't find the drift velocity without knowing it's voltage across it, which I need to know its length to figure out its resistance to figure out its voltage across it by the potential divider principle.