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an n-type silicon sample has a resistivity of 5 w-cm at 300 k.(a) what is the donor impurity concentration?(b) what is the expected resistivity at (

User Picker
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Final answer:

In order to find the donor impurity concentration, use the formula n = 1/(q*R*d), where q is the electron charge, R is the resistivity, and d is the electron mobility. The resistivity at T=200 K is approximately 3.33 Ω·cm, and at T=400 K is approximately 6.67 Ω·cm.

Step-by-step explanation:

To find the donor impurity concentration, we can use the formula:

n = 1/(q*R*d)

where n is the donor impurity concentration, q is the electron charge (1.6 x 10^-19 C), R is the resistivity (5 Ω·cm), and d is the electron mobility (approximately 3000 cm^2/V·s for silicon).

Substituting the given values:

n = 1/(1.6 x 10^-19 C * 5 Ω·cm * 3000 cm^2/V·s)

n ≈ 1.04 x 10^16 impurities/cm^3

To find the resistivity at T=200 K and T=400 K, we can use the formula:

R2 = R1 * (T2 / T1)

where R1 is the resistivity at T=300 K (5 Ω·cm), T1 is the initial temperature (300 K), T2 is the new temperature (200 K or 400 K), and R2 is the new resistivity.

Let's calculate the new resistivities:

  • R2 at T=200 K = 5 Ω·cm * (200 K / 300 K) ≈ 3.33 Ω·cm
  • R2 at T=400 K = 5 Ω·cm * (400 K / 300 K) ≈ 6.67 Ω·cm

Your question is incomplete, but most probably the full question was:

An n-type silicon sample has a resistivity of 5Ω−cm at T=300 K.

a.) What is the donor impurity concentration?

b.) What is the expected resistivity of the material at T=200 K and T=400 K ?

User Oleg Melnikov
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