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The gap between valence and conduction bands in silicon is 1.12$\mathrm { eV } .$ A nickel nucleus in an excited state emits a gamma-ray photon with wavelength $9.31 \times 10 ^ { - 4 } \mathrm { nm }$ . How many electrons can be excited from the top of the valence band to the bottom of the conduction band by the absorption of this gamma ray?

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$1.20 \times 10^{6}$ electrons

Physics 103

Chapter 29

Atoms, Molecules, and Solids

Atomic Physics

Nuclear Physics

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

McMaster University

Lectures

02:42

Atomic physics is the field of physics that studies atoms as an isolated system of electrons and an atomic nucleus. It is primarily concerned with the arrangement of electrons around the nucleus and the processes by which these arrangements change. The theory of quantum mechanics, a set of mathematical rules that describe the behaviour of matter and its interactions, provides a good model for the description of atomic structure and properties.

02:26

In physics, nuclear physics is the field of physics that studies the constituents and interactions of atomic nuclei. The most commonly known applications of nuclear physics are nuclear power generation and nuclear weapons technology, but the research has provided application in many fields, including those in nuclear medicine and magnetic resonance imaging, ion implantation in materials engineering, and radiocarbon dating in geology and archaeology.

03:45

The gap between valence an…

01:19

03:16

00:30

02:03

02:05

The energy gap between val…

00:59

The forbidden gap in silic…

02:18

03:19

(II) The energy gap betwee…

03:24

01:52

A valence electron in a cr…

02:36

The energy gap between the…

00:45

02:09

The band gap in gallium ar…

okay. And this problem we're concerned with how many electrons air gonna get of promoted to the conduction band for silicon. If the gap is 1.12 1.12 you be. And we know the energy given. Um well, we know the wavelength so we can figure out the energy. Um, kind of dumped into the system is HC over lambda. And we're told that Lambda is equal two 9.31 times 10 to the minus four nanometers. So 0.9 point 31 times 10 to the minus 13 meters. Wasn't three water. Okay, great. Those 31 So, um, what we want to do is we want to say, OK, how much total energy do we have? And then we want to figure out how much energy each electron costs and then divide those out to get the number of electrons. And so, if I plug this lambda and to hear I am Tae in, that is equal to um see, um, you just double check my units. Oh, I guess I should use. So I'm gonna read you my copulation. I was thinking about it differently, but I think the way I'm presenting it now. It's hopefully clearer. So the energy now perf Oh Tien is, um I need to convert it to electron rolls. Minus 19 is 2.11 times 10 of the six I'll e v Oops, 10 of the plus six e v. So we want to divide this number by this number to get the total number of electrons. So I guess I'll call this e Gamma. So let's say and the number of Excuse me, electrons is equal to, um, the energy we have divided by the energy per electron. So you gap. So go ahead and plug that into a calculator. It about about my 1.12 Okay, I actually got a different number than I got before, although it's sort of clothes, which is kind of funny. Fine. Um, someone a plus the video while I sort this out. Oh, I realized I was using the reduced Planck's constant. Okay, so now I'm getting what I got earlier, which is, um, 1.19 times 10 to the six electrons

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