Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 2603

In Ge sample, traces of gallium are added as impurity. The resultant sample would behave like: (A) a conductor (B) a P-type semiconductor (C) an N-type semiconductor (D) an insulator

Short Answer

Expert verified
The correct answer is (B) a P-type semiconductor, as adding Gallium (a trivalent impurity) to Germanium creates a P-type semiconductor.

Step by step solution

01

Understanding impurities and their effect on semiconductors

An impurity, also known as dopant, is any foreign material added to a pure semiconductor substance to alter its conductive properties. A small amount of impurity can significantly change the conductive properties of the semiconductor. Impurities can be either trivalent (three valence electrons) or pentavalent (five valence electrons). The addition of a trivalent impurity to a semiconductor results in a P-type semiconductor, and a pentavalent impurity results in an N-type semiconductor.
02

Identify the valency of Gallium and Germanium

To determine the type of semiconductor formed when adding Gallium (Ga) to Germanium (Ge), we must identify their valence electrons. Gallium, having the electron configuration [Ar] 3d^10 4s^2 4p^1, has three valence electrons in its outermost shell. Germanium, having the electron configuration [Ar] 3d^10 4s^2 4p^2, has four valence electrons in its outermost shell.
03

Determine the type of semiconductor

As Gallium has three valence electrons (trivalent), adding it as an impurity to Germanium will create a P-type semiconductor. The extra positive holes caused by Gallium's missing fourth valence electron provide charge carriers for conductivity.
04

Choose the correct answer

From the given options: (A) a conductor (B) a P-type semiconductor (C) an N-type semiconductor (D) an insulator The correct answer is (B) a P-type semiconductor, as adding Gallium (trivalent impurity) to Germanium creates a P-type semiconductor.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A potential barrier of \(0.6 \mathrm{~V}\) exists across a P-N junction. If the depletion region is \(1 \mu \mathrm{m}\) wide, what is the intensity of electric field in the region? (A) \(4 \times 10^{5} \mathrm{Vm}^{-1}\) (B) \(5 \times 10^{5} \mathrm{Vm}^{-1}\) (C) \(6 \times 10^{5} \mathrm{Vm}^{-1}\) (D) \(2 \times 10^{5} \mathrm{Vm}^{-1}\)

The Common Emitter amplifier has voltage gain equal to 300 and its input signal is \(0.5 \cos (100 t)\) volt. The output signal will be equal to (A) \(150 \cos (100 \mathrm{t})\) (B) \(300 \cos (100 \mathrm{t})\) (C) \(150 \cos (100 t+\pi)\) (D) \(300 \cos (100 t-\pi)\)

Digital circuits can be made to be respective use of: (A) AND gate (B) OR gate (C) NOT gate (D) NAND gate

In a common emitter amplifier, output resistance is \(5000 \Omega\) and input resistance is \(1000 \Omega .\) If peak value of signal voltage is $1 \mathrm{mV}\( and \)\beta=100$, then the peak value of output voltage is (A) \(0.1 \mathrm{~V}\) (B) \(0.3 \mathrm{~V}\) (C) \(0.2 \mathrm{~V}\) (D) \(0.5 \mathrm{~V}\)

A n-p-n transistor circuit has \(\alpha=0.985 .\) If $\mathrm{I}_{\mathrm{c}}=9 \mathrm{~mA}\( then the value of \)\mathrm{I}_{\mathrm{b}}$ is (A) \(0.003 \mathrm{~mA}\) (B) \(0.66 \mathrm{~mA}\) (C) \(0.015 \mathrm{~mA}\) (D) \(0.13 \mathrm{~mA}\)
See all solutions

Recommended explanations on English Textbooks

International English

Read Explanation

Blog

Read Explanation

Pragmatics

Read Explanation

Global English

Read Explanation

English Language Study

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free