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Chapter 10: Problem 63

As a crude approximation, an impurity pentavalent atom in a (tetravalent) silicon lattice can be treated as a one-electron atom, in which the extra electron orbits a net positive charge of 1 . Because this "atom" is not in free space, however, the permitivity of free space, \(\varepsilon_{0}\) must be replaced by \(\kappa \varepsilon_{0}\), where \(\kappa\) is the dielectric constant of the surrounding material. The hydrogen atom ground-state energies would thus become $$ E=-\frac{m e^{4}}{2\left(4 \pi \kappa \varepsilon_{0}\right)^{2} h^{2}} \frac{1}{n^{2}}=\frac{-13.6 \mathrm{eV}}{\kappa^{2} n^{2}} $$ Given \(\kappa=12\) for silicon, how much ener gy is needed to frec a donor electron in its ground state? (Actually. the effective mass of the donor electron is less than \(m_{e}\), so this prediction is somewhat high.)

Short Answer

Expert verified
The energy needed to free a donor electron in its ground state in silicon is approximately 0.0944 eV.

Step by step solution

01

Understand the variables in the equation

Here, the formula for energy is given as \(E=-\frac{13.6 eV}{\kappa^{2} n^{2}}\). Here, n is the number that denotes the state of the electron (n=1 for ground state), eV is a unit of energy called electronvolt, and 魏 is the dielectric constant of the material (silicon), which is given as 12 in this case.
02

Substitute values into the equation

In the energy equation, substitute the given values: \(n = 1\) (since it's the ground state) and \(\kappa = 12\), we get: \(E=-\frac{13.6 eV}{(12)^{2} (1)^{2}}\).
03

Simplify the equation to find the answer

Simplify the expression to calculate the energy. This will yield: \(E=-\frac{13.6 eV}{144} = -0.0944 eV\). The negative sign indicates that the energy is needed to take the electron from this state, i.e., the energy must be supplied to the system. Thus, the absolute value will be considered to get the energy required.

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Key Concepts

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

Understanding Dielectric Constant
The dielectric constant, often represented by the symbol \(\kappa\), is a crucial parameter in semiconductor physics. It reflects a material's ability to screen electric fields, which affects how charges interact within the material. In simple terms, dielectric constant is a measure of how much a material can reduce the effective force between two charges as compared to the force experienced in a vacuum.

When we discuss the dielectric constant in the context of semiconductors, like silicon in the given exercise, it significantly affects the behavior of charge carriers such as electrons. Silicon's high dielectric constant of 12 implies that the Coulomb forces, which would ordinarily bind an electron to a nucleus very tightly in a vacuum, are instead substantially screened. This decreased force allows for an electron to be more loosely bound in the solid, affecting its energy levels and how much energy is needed to free it, as seen in the exercise.
Energy Levels in Semiconductors
Energy levels in semiconductors are crucial to their electronic properties. Unlike conductors, semiconductors have a band gap between the valence band (filled with electrons) and conduction band (where electrons can move freely). The energy levels of electrons determine their position in these bands.

Impurities, such as pentavalent atoms in silicon, introduce additional energy levels known as donor or acceptor levels. These levels are typically close to the conduction or valence band, respectively, and thus require much less energy to free an electron (or to move a hole) as compared to intrinsic silicon. The exercise demonstrates the calculation of this energy for an electron in a donor impurity level, which is significantly lower than the energy needed to lift an electron across the silicon's band gap due to the higher dielectric constant.
Effective Mass of an Electron
The concept of effective mass arises from the fact that an electron moving through a crystalline lattice behaves differently than it would in free space. The periodic potential of the lattice affects the electron's motion, and as a result, its mass appears to differ from the mass of a free electron.

In the context of our exercise, the effective mass of the electron is mentioned as being less than the electron mass in free space (\(m_e\)). This reduction in effective mass reflects an enhanced response to external forces due to the interaction with the periodic potential of the lattice. A lower effective mass means it鈥檚 easier to move the electron within the crystal lattice, directly affecting mobility and conductivity. Consequently, the energy calculation from the exercise uses a value that is slightly higher than the real case, since it doesn't take into account the lower effective mass.

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Most popular questions from this chapter

The magnetic field at the surface of a long wire of radius \(R\) and cairying a current \(I\) is \(\mu_{0} I / 2 \pi R\). How large a curtent could a 0.1 mm diameter niobium wire carry without exceeding its \(0.2 \mathrm{~T}\) critical field?

What is a Cooper pair, and what role does it play in superconductivity?

The "floating magnet trick" is shown in Figure \(10.50 .\) If the disk on the bottom were a permanent magnet, rather than a superconductor, the trick wouldn't work. The superconductor does produce un external field very similar to that of a permanent magnet. What other characteristic is necessary to explain the effect? (Hint: What happens when you hold two ordinary magnets so that they repel, and then you release one of them?)

Electron affinity is a property specif ying the "appetite" of an element for gaining electrons. Elements, such as fluorine and oxygen, that lack only one or two electrons to complete shells can achieve a lower energy state by absorbing an external electron. For instance, in uniting an electron with a neutral chlorine atom. completing its \(n=3\) shell and forming a \(\mathrm{Cl}^{-}\) ion, \(3.61 \mathrm{eV}\) of energy is liberated. Suppose an electron is detached from a sodium atom, whose ionization energy is \(5.14 \mathrm{eV}\), then transferred to a (faraway) chlorine atom. (a) Must energy on balance be put in by an external agent, or is some energy actually liberated? If so.how much" (b) The transfer leaves the sodium with a positive charge and the chlorine with a negative. Energy can now be extracted by allowing these ions to draw close, forming a molecule. How close must they approach to rccmer the energy crpended in part (a)? (c) The actual seperstion of the atoms in a NaCI molecule is \(0.24 \mathrm{nm}\). How much lower in energy is the molecule than the icparated neutral atums?

Describe the similarities and differences between Type-l and Type-Il superconductors.
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