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Chapter 18: Problem 29

(a) The room-temperature electrical conductivity of a silicon specimen is \(500(\Omega-\mathrm{m})^{-1}\) The hole concentration is known to be \(2.0 \times 10^{22} \mathrm{m}^{-3}\). Using the electron and hole mobilities for silicon in Table \(18.3,\) compute the electron concentration. (b) On the basis of the result in part (a), is the specimen intrinsic \(n\) -type extrinsic, or \(p\) -type extrinsic? Why?

Short Answer

Expert verified
Answer: The electron concentration is approximately \(5.0 \times 10^{22} \mathrm{m}^{-3}\), and the specimen is n-type extrinsic.

Step by step solution

01

Write down the given values and formula for electrical conductivity

Given: Electrical conductivity (\(\sigma\)) = \(500(\Omega-\mathrm{m})^{-1}\) Hole concentration (p) = \(2.0 \times 10^{22} \mathrm{m}^{-3}\) Electron mobility (\(\mu_n\)) for silicon = \(0.13\ \mathrm{m}^2\mathrm{V}^{-1}\mathrm{s}^{-1}\) Hole mobility (\(\mu_p\)) for silicon = \(0.05\ \mathrm{m}^2\mathrm{V}^{-1}\mathrm{s}^{-1}\) The formula for electrical conductivity is: \(\sigma = q(n\mu_n + p\mu_p)\)
02

Solve the formula for electron concentration (n)

We need to find the electron concentration (n). The formula can be rearranged to solve for n as follows: \(n = \frac{\sigma - q(p\mu_p)}{q\mu_n}\)
03

Calculate the electron concentration

Now, substitute the given values and constants into the formula: \(q = 1.6 \times 10^{-19} \mathrm{C}\) \(\sigma = 500 (\Omega-\mathrm{m})^{-1}\) \(p = 2.0 \times 10^{22} \mathrm{m}^{-3}\) \(\mu_n = 0.13\ \mathrm{m}^2\mathrm{V}^{-1}\mathrm{s}^{-1}\) \(\mu_p = 0.05\ \mathrm{m}^2\mathrm{V}^{-1}\mathrm{s}^{-1}\) \(n = \frac{500 - 1.6 \times 10^{-19}(2.0 \times 10^{22} \times 0.05)}{1.6 \times 10^{-19} \times 0.13}\) \(n \approx 5.0 \times 10^{22} \mathrm{m}^{-3}\) The electron concentration is approximately \(5.0 \times 10^{22} \mathrm{m}^{-3}\).
04

Determine the type of specimen based on electron concentration

Now that we have found the electron concentration (which is two times the hole concentration), we can determine whether the specimen is intrinsic, n-type extrinsic, or p-type extrinsic. Since the electron concentration (n) is greater than the hole concentration (p), the specimen is n-type extrinsic. In an n-type extrinsic semiconductor, the majority charge carriers are electrons, and the minority charge carriers are holes.

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Most popular questions from this chapter

How does the clectron structure of an isolated atom differ from that of a solid material?

In terms of electron energy band structure, discuss reasons for the difference in electrical conductivity between metals, semiconductors, and insulators

Compare the temperature dependence of the conductivity for metals and intrinsic semiconductors. Briefly explain the difference in behavior.

Some metal alloy is known to have electrical conductivity and electron mobility values of \(1.2 \times 10^{7}(\Omega-\mathrm{m})^{-1}\) and \(0.0050 \mathrm{m}^{2} / \mathrm{N}-\mathrm{s}\), respec- tively. Through a specimen of this alloy that is \(35 \mathrm{mm}\) thick is passed a current of \(40 \mathrm{A}\). What magnetic field would need to be imposed to yield a Hall voltage of \(-3.5 \times 10^{-7} \mathrm{V} ?\)

An \(n\) -type semiconductor is known to have an electron concentration of \(5 \times 10^{17} \mathrm{m}^{-3}\). If the electron drift velocity is \(350 \mathrm{m} / \mathrm{s}\) in an electric field of \(1000 \mathrm{V} / \mathrm{m}\), calculate the conductivity of this material
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