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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?

Short Answer

Expert verified
So, the absorption of this gamma ray can excite approximately \(1.19 \times 10^{5}\) electrons.

Step by step solution

01

Convert Energy from eV to Joules

The energy required to excite an electron from the valence band to the conduction band is given as \(1.12 eV\). We know that \(1 eV = 1.6 \times 10^{-19} J\). Therefore, the energy required in Joules would be \(1.12 eV \times 1.6 \times 10^{-19} J/eV = 1.79 \times 10^{-19} J\)
02

Calculate Energy of the Gamma Ray

The energy of the gamma-ray photon can be calculated with the formula \(E = \frac{hc}{\lambda}\), where \(h = 6.63 \times 10^{-34} Js\) is Planck's constant, \(c = 3.0 \times 10^{8} m/s\) is the speed of light, and \( \lambda =9.31 \times 10^{-12} m\) is the wavelength. So, \(E = \frac{(6.63 \times 10^{-34} Js)(3.0 \times 10^{8} m/s)}{9.31 \times 10^{-12} m} = 2.14 \times 10^{-14} J\)
03

Determine the Number of Electrons

To find the number of electrons that can be excited by the absorption of this gamma ray, we need to divide the energy of the gamma ray by the energy required by each electron. So the number of electrons = \(\frac{2.14 \times 10^{-14} J}{1.79 \times 10^{-19} J} = 1.19 \times 10^{5}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Band Gap in Semiconductors
The concept of a band gap is central to semiconductor physics and the functioning of various electronic devices. In a semiconductor like silicon, the band gap represents the energy difference between the valence band, which is filled with electrons, and the conduction band, where electrons can move freely and conduct electricity.

In simple terms, the band gap is a sort of 'no man’s land' for electrons within an atom where no electron states can exist. It's like an energy barrier that electrons need to overcome to move from being bound (in the valence band) to being free to conduct current (in the conduction band). For silicon, this band gap is about 1.12 electron volts (eV).

It is crucial for students to understand that the size of the band gap determines a material's electrical properties. A large band gap results in an insulator, a moderate gap in a semiconductor, and a very small gap leads to a conductor. The band gap is an intrinsic property of the material and is of high significance in determining which photons can be absorbed to excite electrons.
Valence to Conduction Band Transition
When discussing semiconductors, one often hears about the transition of electrons from the valence to the conduction band. In essence, this transition is the leap an electron makes when it absorbs enough energy to overcome the band gap.

An everyday analogy could be jumping from one ledge to a higher one; the energy you need to make the jump would represent the band gap energy. That’s why in our silicon example, an electron requires an energy of at least 1.12 eV to make the transition. Once electrons gain this energy, which can come from heat, light, or electrical fields, they can contribute to electrical current.

Understanding this concept is critical when dealing with phenomena such as photoelectric effect or semiconductor light absorption — it's all about giving electrons enough energy to make the jump to a higher state, which is exactly what happens with our gamma-ray photon in the exercise.
Gamma-Ray Photon Absorption
Gamma-ray photon absorption is an intriguing phenomenon that occurs at the atomic level. Gamma rays, which are high-energy photons, carry enough energy to excite electrons from the valence band across the band gap into the conduction band.

Let's unpack this using the given problem as an example. A gamma-ray photon emitted from an excited nickel nucleus has a tiny wavelength, implying very high energy (as energy and wavelength are inversely proportional). When such a photon is absorbed by a silicon atom, its energy is transferred to an electron in the valence band. If this energy exceeds the band gap of 1.12 eV, the electron can be excited to the conduction band, contributing to the material's conductivity.

The exercise provided requires calculating the number of electrons that can be excited by a single gamma-ray photon. We use fundamental constants like Planck’s constant and the speed of light to find the energy of the photon and then see how many electronic band gap jumps that energy accounts for—a powerful concept linking quantum mechanics to practical electronic applications.

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Most popular questions from this chapter

A hypothetical NH molecule makes a rotational-level transition from \(l=3\) to \(l=1\) and gives off a photon of wavelength \(1.780 \mathrm{nm}\) in doing so. What is the separation between the two atoms in this molecule if we model them as point masses? The mass of hydrogen is \(1.67 \times 10^{-27} \mathrm{~kg},\) and the mass of nitrogen is \(2.33 \times 10^{-26} \mathrm{~kg}\)

Two atoms of cesium (Cs) can form a Cs \(_{2}\) molecule. The equilibrium distance between the nuclei in a \(\mathrm{Cs}_{2}\) molecule is \(0.447 \mathrm{nm}\). Calculate the moment of inertia about an axis through the center of mass of the two nuclei and perpendicular to the line joining them. The mass of a cesium atom is \(2.21 \times 10^{-25} \mathrm{~kg}\).

When an OH molecule undergoes a transition from the \(n=0\) to the \(n=1\) vibrational level, its internal vibrational energy increases by \(0.463 \mathrm{eV}\). Calculate the frequency of vibration and the force constant for the interatomic force. (The mass of an oxygen atom is \(2.66 \times 10^{-26} \mathrm{~kg},\) and the mass of a hydrogen atom is \(1.67 \times 10^{-27} \mathrm{~kg} .\)

In the discussion of free electrons in Section \(42.5,\) we assumed that we could ignore the effects of relativity. This is not a safe assumption if the Fermi energy is greater than about \(\frac{1}{100} m c^{2}\) (that is, more than about \(1 \%\) of the rest energy of an electron). (a) Assume that the Fermi energy at absolute zero, as given by Eq. (42.19), is equal to \(\frac{1}{100} m c^{2} .\) Show that the electron concentration is $$ \frac{N}{V}=\frac{2^{3 / 2} m^{3} c^{3}}{3000 \pi^{2} \hbar^{3}} $$and determine the numerical value of \(N / V .\) (b) Is it a good approximation to ignore relativistic effects for electrons in a metal such as copper, for which the electron concentration is \(8.45 \times 10^{28} \mathrm{~m}^{-3}\) ? Explain. (c) A white dwarf star is what is left behind by a star like the sun after it has ceased to produce energy by nuclear reactions. (Our own sun will become a white dwarf star in another \(8 \times 10^{9}\) years or so.) A typical white dwarf has mass \(2 \times 10^{30} \mathrm{~kg}\) (comparable to the sun) and radius \(6000 \mathrm{~km}\) (comparable to that of the earth). The gravitational attraction of different parts of the white dwarf for each other tends to compress the star; what prevents it from compressing is the pressure of free electrons within the star (see Challenge Problem 42.53 ). Use both of the following assumptions to estimate the electron concentration within a typical white dwarf star: (i) the white dwarf star is made of carbon, which has a mass per atom of \(1.99 \times 10^{-26} \mathrm{~kg} ;\) and (ii) all six of the electrons from each carbon atom are able to move freely throughout the star. (d) Is it a good approximation to ignore relativistic effects in the structure of a white dwarf star? Explain.

During each of these processes, a photon of light is given up. In each process, what wavelength of light is given up, and in what part of the electromagnetic spectrum is that wavelength? (a) A molecule decreases its vibrational energy by \(0.198 \mathrm{eV} ;\) (b) an atom decreases its energy by \(7.80 \mathrm{eV} ;\) (c) a molecule decreases its rotational energy by \(4.80 \times 10^{-3} \mathrm{eV}\)

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