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Chapter 5: Problem 38

Refer to a particle of mass \(m\) trapped in a half-infinite well, with potential energy given by $$ U(x)=\left\\{\begin{array}{ll} \infty & x \leq 0 \\\0 & 0< x< L \\\U_{0} & x \geq L\end{array}\right.$$ Advance an argument based on \(p=h / \lambda\) that there is \(\mathrm{po}\) bound state in a half -infinite well unless \(U_{0}\) is at least \(h^{2} / 32 m L^{2}\). (Hint: What is the maximum wavelength possible within the well?)

Short Answer

Expert verified
For the bound state to exist in a half-infinite well, the condition is that \(U_{0}\) must be at least equal to \( h^{2} / 8mL^{2} \)

Step by step solution

01

Find the maximum wavelength within the well

With the particle trapped in the well from 0 to \(L\), the longest wavelength permissible would be equivalent to twice the length of the well. This is because the well has to contain at least half a wavelength, a necessity for a standing wave which is implicitly required for a bound state in quantum mechanics. Therefore, \( \lambda_{\text{max}} = 2L \).
02

Utilize the de Broglie relation to find the minimum momentum

We know from the de Broglie's principle that the momentum and wavelength are inversely proportional: \( p = h / \lambda \). By substituting the value of \( \lambda_{\text{max}} \) from the previous step we find that \( p_{\text{min}} = h / 2L \). The particle must have at least this momentum to be in a bound state.
03

Derive the minimum kinetic energy

We know that for a particle, the kinetic energy (\(K.E.\)) is given by \( p^{2}/2m \). Substituting the value for \( p_{\text{min}} \) yields a minimum kinetic energy of \( h^{2} / 8mL^{2} \).
04

Find the condition for the bound state

In order for the particle to be in a bound state, the total energy should be less than \(U_{0}\). The total energy is the sum of the potential and kinetic energy. For the half-infinite well, the total energy is therefore \( K.E. + U_{0} \), or \( h^{2} / 8mL^{2} + U_{0} \). For a bound state, this must be less than \(U_{0}\) yielding the inequality \(U_{0}\geq h^{2}/8mL^{2}\). This inequality sets the lower limit for \(U_{0}\) in order for a bound state to exist.

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Key Concepts

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

Quantum Well
A quantum well is like a tiny pocket that looks almost infinite. Here, a particle, like an electron, is trapped and can't escape easily. Imagine holding a marble in a narrow trough, bouncing back and forth. This reflects the concept of a quantum well, where potential energy restricts the particle's movement to a defined region.
In a half-infinite well, one side is infinitely high, keeping the particle from crossing over.
  • The shape of the well determines how the particle behaves inside it.
  • The trap allows for energy levels, which are specific states a particle can indulge in.
  • The energy levels are quantized, meaning not every energy level is permissible.
The well in the exercise starts at zero potential energy within it, with infinity making the escape impossible on one side, and a finite barrier on the other keeping it inside easier than crawling out.
de Broglie Wavelength
The de Broglie wavelength relates to how we view particles as waves. This is because, in quantum mechanics, particles exhibit wave-like properties.
In our problem, the particle inside the well has a longest possible wavelength for the lowest energy state. This longest possible wave fits perfectly within the well, like drawing a wave between two ends of a string that's fixed at both ends. The formula connecting wavelength and momentum is \[ p = \frac{h}{\lambda} \]where:
  • \(h\) is the Planck's constant, a fundamental value in quantum mechanics.
  • \(\lambda\) is the de Broglie wavelength of the particle inside the well.
The relation shows that a longer wavelength means less momentum for the particle, crucial for determining the conditions for a bound state in a quantum well.
Momentum-Energy Relation
Momentum relates directly to energy for a particle inside a quantum well. When discussing quantum mechanics, a particle's energy in a specific state can be found knowing its momentum. The concept utilizes the formula where kinetic energy is expressed as:\[ K.E. = \frac{p^2}{2m} \]Where:
  • \(p\) indicates the momentum needing to be calculated first.
  • \(m\) represents the particle's mass, which is constant in our setup.
After determining the minimum momentum using \( p = \frac{h}{2L} \), one calculates kinetic energy. This minimal kinetic energy is important. It sets a benchmark to compare against the well's potential energy barrier (\(U_0\)). Knowing both momentum and energy allows for defining if a particle can be bound or has enough energy to escape.
Bound State
A bound state in quantum mechanics is when a particle remains trapped within a potential well, unable to escape due to insufficient energy. The energy required to maintain a bound state within the quantum well must be lower than the barrier height potential \(U_0\). For this particular exercise, the solution uses momentum and energy calculation to find:\[ U_0 \geq \frac{h^2}{8mL^2} \]Here:
  • The kinetic energy is found by minimizing the wavelength and momentum.
  • The potential energy \(U_0\) is the barrier level the particle faces at the edge of the well.
If the particle's total energy ('kinetic + potential') remains less than \(U_0\), it stays bound, otherwise, it may transition out of the well into a different energy state or spatial position.

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Most popular questions from this chapter

Harmonic Oscillator: The hannonic oscillator can be solved exactly for the quantized energies, and here we compare those results with a numerical approach. Along with the values of \(m\) and \(\hbar\) discussed above, we choose our units so that the spring constant \(x\) is also 1 . The potential energy function \(U(x)\) is then simply \(\frac{1}{2} x^{2}\) For \(\Delta x\), use \(0.001\). (a) Following the above guidelines on choosing \(1 \beta(0)\) and \(\psi(\Delta x)\), test both odd and even functions at different trial values of \(E\) by finding \(\psi\) at all positive multiples of \(\Delta x\) out \(10 x=4\) and plotting the results. Note that because of the functions' symmetries there is no need to plot negative values of \(x\). Find four allowed energies. (b) What tells you that an energy is correct? (c) Compare your results with the exact values given in equation \((5-26)\).

In a study of heat transfer, we find that for a solid rod. there is a relationship between the second derivative of the temperature with respect to position along the rod and the first with respect to time. (A linear temperature change with position would imply as much heat flowing into a region as out. so the temperature there would not change with time.) $$ \frac{\partial^{2} T(x, t)}{\partial x^{2}}=b \frac{\partial T(x, t)}{\partial t} $$ (a) Separate variables. That is, assume a solution that is a product of a function of \(x\) and a function of \(1 .\) plug it in. then divide by it. Obtain two ordinary differential equations. (b) Consider a fairly simple, if somewlat unrealistic. case. Suppose the temperature is 0 at \(x=0\) and \(x=L\), and positive in betwaen. Write down the simplest function of.r that (1) fits these conditions and \((2)\) obey s the differential equation involving x. Does your choice determine the value. including sign. of some constant? (c) Obtain the full \(T(x, t)\) for this case.

A study of classical waves tells us that a standing wave can be expressed as a sum of two traveling waves. Quantum-mechanical traveling waves, discussed in Chapter \(4,\) are of the form \(\Psi(x, t)=A e^{i(h x-\alpha t)} .\) Show that the infinite well's standing-wave function can be expressed as a sum of two traveling waves.

Where would a particle in the fist excited state (first above ground) of an infinite well mostly likely be found?

A student of classical physics says, "A charged particle. like an electron orbiting in u simple atom. shouldn't have only certain stable energies: in fact, it should lose energy by electromagnetic radiation until the atom collapses." Answer these two complaints qualitatively. appealing to as few fundumentul claims of quantum mechanics as possible
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