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Chapter 40: Problem 43
Consider a beam of free particles that move with velocity \(v=p / m\) in the \(x\)
-direction and are incident on a potential-energy step \(U(x)=0,\) for \(x<0,\)
and \(U(x)=U_{0}
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When low-energy electrons pass through an ionized gas, electrons of certain energies pass through the gas as if the gas atoms weren't there and thus have transmission coefficients (tunneling probabilities) \(T\) equal to unity. The gas ions can be modeled approximately as a rectangular barrier. The value of \(T=1\) occurs when an integral or half-integral number of de Broglie wavelengths of the electron as it passes over the barrier equal the width \(L\) of the barrier. You are planning an experiment to measure this effect. To assist you in designing the necessary apparatus, you estimate the electron energies \(E\) that will result in \(T=1\). You assume a barrier height of \(10 \mathrm{eV}\) and a width of \(1.8 \times 10^{-10} \mathrm{~m} .\) Calculate the three lowest values of \(E\) for which \(T=1\)
Consider a potential well defined as \(U(x)=\infty\) for \(x<0, U(x)=0 \quad\) for
\(\quad 0
A particle of mass \(m\) in a one-dimensional box has the following wave function in the region \(x=0\) to \(x=L:\) $$\Psi(x, t)=\frac{1}{\sqrt{2}} \psi_{1}(x) e^{-i E_{1} t / \hbar}+\frac{1}{\sqrt{2}} \psi_{3}(x) e^{-i E_{3} t / \hbar}$$ Here \(\psi_{1}(x)\) and \(\psi_{3}(x)\) are the normalized stationary-state wave functions for the \(n=1\) and \(n=3\) levels, and \(E_{1}\) and \(E_{3}\) are the energies of these levels. The wave function is zero for \(x<0\) and for \(x>L\) (a) Find the value of the probability distribution function at \(x=L / 2\) as a function of time. (b) Find the angular frequency at which the probability distribution function oscillates.
An electron with initial kinetic energy \(6.0 \mathrm{eV}\) encounters a barrier with height \(11.0 \mathrm{eV}\). What is the probability of tunneling if the width of the barrier is (a) \(0.80 \mathrm{nm}\) and (b) \(0.40 \mathrm{nm} ?\)
The WKB Approximation. It can be a challenge to solve the Schr枚dinger equation
for the bound-state energy levels of an arbitrary potential well. An
alternative approach that can yield good approximate results for the energy
levels is the \(W K B\) approximation (named for the physicists Gregor Wentzel,
Hendrik Kramers, and L茅on Brillouin, who pioneered its application to quantum
mechanics). The WKB approximation begins from three physical statements: (i)
According to de Broglie, the magnitude of momentum \(p\) of a quantum-mechanical
particle is \(p=h / \lambda\). (ii) The magnitude of momentum is related to the
kinetic energy \(K\) by the relationship \(K=p^{2} / 2 m .\) (iii) If there are no
nonconservative forces, then in Newtonian mechanics the energy \(E\) for a
particle is constant and equal at each point to the sum of the kinetic and
potential energies at that point: \(E=K+U(x),\) where \(x\) is the coordinate. (a)
Combine these three relationships to show that the wavelength of the particle
at a coordinate \(x\) can be written as
$$ \lambda(x)=\frac{h}{\sqrt{2 m[E-U(x)]}} $$ Thus we envision a quantum-
mechanical particle in a potential well \(U(x)\) as being like a free particle,
but with a wavelength \(\lambda(x)\) that is a function of position. (b) When
the particle moves into a region of increasing potential energy, what happens
to its wavelength?
(c) At a point where \(E=U(x),\) Newtonian mechanics says that the particle has
zero kinetic energy and must be instantaneously at rest. Such a point is
called a classical turning point, since this is where a Newtonian particle
must stop its motion and reverse direction. As an example, an object
oscillating in simple harmonic motion with amplitude \(A\) moves back and forth
between the points \(x=-A\) and \(x=+A ;\) each of these is a classical turning
point, since there the potential energy \(\frac{1}{2} k^{\prime} x^{2}\) equals
the total energy \(\frac{1}{2} k^{\prime} A^{2}\). In the WKB expression for
\(\lambda(x),\) what is the wavelength at a classical turning point? (d) For a
particle in a box with length \(L,\) the walls of the box are classical turning
points (see Fig. 40.8\()\) Furthermore, the number of wavelengths that fit
within the box must be a half-integer (see Fig. 40.10 ), so that \(L=(n / 2)
\lambda\) and hence \(L / \lambda=n / 2,\) where \(n=1,2,3, \ldots\) [Note that
this is a restatement of Eq. (40.29).] The WKB scheme for finding the allowed
bound-state energy levels of an arbitrary potential well is an extension of
these observations. It demands that for an allowed energy \(E\), there must be a
half-integer number of wavelengths between the classical turning points for
that energy. Since the wavelength in the WKB approximation is not a constant
but depends on \(x\), the number of wavelengths between the classical turning
points \(a\) and \(b\) for a given value of the energy is the integral of \(1 /
\lambda(x)\) between those points: $$ \int_{a}^{b} \frac{d
x}{\lambda(x)}=\frac{n}{2} \quad(n=1,2,3, \ldots) $$
Using the expression for \(\lambda(x)\) you found in part (a), show that the \(W
K B\) condition for an allowed bound-state energy can be written as $$
\int_{a}^{b} \sqrt{2 m[E-U(x)]} d x=\frac{n h}{2} \quad(n=1,2,3, \ldots) $$
(e) As a check on the expression in part (d), apply it to a particle in a box
with walls at \(x=0\) and \(x=L\). Evaluate the integral and show that the allowed
energy levels according to the WKB approximation are the same as those given
by Eq. (40.31). (Hint: since the walls of the box are infinitely high, the
points \(x=0\) and \(x=L\) are classical turning points for any energy \(E .\)
Inside the box, the potential energy is zero.) (f) For the finite square well
shown in Fig. \(40.13,\) show that the \(\mathrm{WKB}\) expression given in part
(d) predicts the same bound-state energies as for an infinite square well of
the same width. (Hint: Assume \(E
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