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A proton with initial kinetic energy 50.0 eV encounters a barrier of height 70.0 eV. What is the width of the barrier if the probability of tunneling is 8.0 \(\times\) 10\(^{-3}\)? How does this compare with the barrier width for an electron with the same energy tunneling through a barrier of the same height with the same probability?

Short Answer

Expert verified
The proton barrier width is \( 5.01 \times 10^{-11} \text{ m} \) and the electron's is \( 3.56 \times 10^{-11} \text{ m} \). The electron's width is smaller.

Step by step solution

01

Understanding Tunneling

Quantum tunneling occurs when a particle passes through a potential barrier that it classically shouldn't be able to surmount. The probability of tunneling is given by the formula \( T = e^{-2 \gamma a} \), where \( \gamma = \sqrt{\frac{2m(U-E)}{\hbar^2}} \), \( U \) is the barrier height, \( E \) is the kinetic energy of the particle, \( m \) is the mass of the particle, and \( a \) is the barrier width.
02

Calculate \( \gamma \) for the Proton

Given the proton's initial kinetic energy is 50.0 eV and the barrier height is 70.0 eV, the energy difference \( U - E = 70.0 - 50.0 = 20.0 \) eV. First, convert this energy to joules. Use \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \). Thus, \( U-E = 20.0 \times 1.602 \times 10^{-19} = 3.204 \times 10^{-18} \, \text{J} \). Now, using \( m_p = 1.67 \times 10^{-27} \, \text{kg} \) for the proton mass and \( \hbar = 1.054 \times 10^{-34} \, \text{J s} \), compute \( \gamma \):\[ \gamma = \sqrt{\frac{2 \times 1.67 \times 10^{-27} \times 3.204 \times 10^{-18}}{(1.054 \times 10^{-34})^2}} = 1.146 \times 10^{10} \, \text{m}^{-1} \]
03

Solve for the Barrier Width \( a \) for the Proton

Using \( T = e^{-2 \gamma a} = 8.0 \times 10^{-3} \), solve for \( a \):\[ e^{-2 \times 1.146 \times 10^{10} a} = 8.0 \times 10^{-3} \]Taking the natural logarithm of both sides gives:\[ -2 \times 1.146 \times 10^{10} a = \ln(8.0 \times 10^{-3}) \]\[ a = \frac{\ln(8.0 \times 10^{-3})}{-2 \times 1.146 \times 10^{10}} \approx 5.01 \times 10^{-11} \text{ m} \]
04

Calculate \( \gamma \) for the Electron

For an electron, use the same energy difference \( U - E = 70.0 - 50.0 = 20.0 \text{ eV} = 3.204 \times 10^{-18} \text{ J} \). Use the electron mass \( m_e = 9.11 \times 10^{-31} \text{ kg} \). Calculate \( \gamma \):\[ \gamma = \sqrt{\frac{2 \times 9.11 \times 10^{-31} \times 3.204 \times 10^{-18}}{(1.054 \times 10^{-34})^2}} = 1.612 \times 10^{10} \text{ m}^{-1} \]
05

Solve for the Barrier Width \( a \) for the Electron

Using \( T = e^{-2 \gamma a} = 8.0 \times 10^{-3} \), solve for \( a \) for the electron:\[ e^{-2 \times 1.612 \times 10^{10} a} = 8.0 \times 10^{-3} \]Taking the natural logarithm of both sides gives:\[ -2 \times 1.612 \times 10^{10} a = \ln(8.0 \times 10^{-3}) \]\[ a = \frac{\ln(8.0 \times 10^{-3})}{-2 \times 1.612 \times 10^{10}} \approx 3.56 \times 10^{-11} \text{ m} \]
06

Compare Proton and Electron Barrier Widths

The calculated barrier width for the proton is approximately \( 5.01 \times 10^{-11} \text{ m} \), while for the electron it is approximately \( 3.56 \times 10^{-11} \text{ m} \). The electron's barrier width is smaller due to its smaller mass, resulting in a larger \( \gamma \) and thus a smaller width for the same tunneling probability.

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Key Concepts

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

Potential Barrier
In the world of quantum mechanics, a potential barrier is like an invisible wall that particles encounter. Imagine it as a hurdle that particles need to jump over. However, in quantum mechanics, things get a bit strange. Particles have a chance to pass through this barrier even if they don't have enough energy to climb over it. This phenomenon is called "quantum tunneling." Unlike classical physics, where an object would need to have energy greater than or equal to the height of the barrier to cross it, quantum mechanics allows particles to pass through barriers based on probability. This is a fascinating concept that shows how particles behave more like waves.
This tunneling effect leads to the formula:
  • \( T = e^{-2 \gamma a} \) where \( T \) is the tunneling probability,
  • \( \gamma = \sqrt{\frac{2m(U-E)}{\hbar^2}} \)
Here, \( m \) is the mass of the particle, \( U \) is the potential energy of the barrier, \( E \) is the particle's energy, and \( a \) is the barrier width. The tunneling probability \( T \) decreases exponentially with an increase in the width of the barrier or the barrier height relative to the energy of the particle.
Kinetic Energy
Kinetic energy is the energy a particle possesses because of its motion. In the context of quantum tunneling, this is the energy that a particle carries as it approaches a potential barrier. The height of the barrier is critical because it must be higher than the particle’s kinetic energy for tunneling to occur.
When a proton or an electron encounters a barrier with more energy than its kinetic energy, the scenario becomes interesting. Even if the particle doesn’t have enough kinetic energy to overcome the barrier classically, there is still a probability, albeit small, that it can tunnel through. This is because in quantum mechanics, phenomena do not require enough energy to pass a barrier but can instead "borrow" energy temporarily, as long it is within the limitations specified by Heisenberg's uncertainty principle.
Think about this energy borrowing as a particle waving energetically at the barrier and sometimes sneaking through due to its wave-like properties. The difference between the particle's kinetic energy \( E \) and the barrier energy \( U \) \( (U - E) \) plays a crucial role in determining the tunneling probability.
Proton and Electron Comparison
Comparing a proton and an electron in quantum tunneling illustrates the effect of a particle's mass on its ability to tunnel through a barrier. Protons are much heavier than electrons. This significant difference in mass influences the tunneling probability due to the equation:
  • \( \gamma = \sqrt{\frac{2m(U-E)}{\hbar^2}} \)
For a proton with a larger mass (\( m_p = 1.67 \times 10^{-27} \text{ kg} \)), \( \gamma \) is smaller compared to an electron (\( m_e = 9.11 \times 10^{-31} \text{ kg} \)), given the same energy difference \( U - E \).
This results in:
  • A smaller barrier width \( a \) for the electron, since its \( \gamma \) is larger, making it more likely to tunnel through a barrier of the same height as compared to a proton.
  • Thus, the required barrier width for a proton is larger (e.g., \( 5.01 \times 10^{-11} \text{ m} \)) compared to an electron (\( 3.56 \times 10^{-11} \text{ m} \)) for the same tunneling probability.
The electron's smaller mass and larger tunneling probability highlights how different masses influence quantum tunneling, with lighter particles more readily penetrating potential barriers than heavier ones for identical conditions.

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Most popular questions from this chapter

In your research on new solid-state devices, you are studying a solid-state structure that can be modeled accurately as an electron in a one-dimensional infinite potential well (box) of width \(L\). In one of your experiments, electromagnetic radiation is absorbed in transitions in which the initial state is the \(n\) = 1 ground state. You measure that light of frequency \(f\) = 9.0 \(\times\) 10\(^{14}\) Hz is absorbed and that the next higher absorbed frequency is 16.9 \(\times\) 10\(^{14}\) Hz. (a) What is quantum number \(n\) for the final state in each of the transitions that leads to the absorption of photons of these frequencies? (b) What is the width \(L\) of the potential well? (c) What is the longest wavelength in air of light that can be absorbed by an electron if it is initially in the \(n\) = 1 state?

A harmonic oscillator consists of a 0.020-kg mass on a spring. The oscillation frequency is 1.50 Hz, and the mass has a speed of 0.480 m/s as it passes the equilibrium position. (a) What is the value of the quantum number n for its energy level? (b) What is the difference in energy between the levels \(E_n\) and \(E_{n+1}\)? Is this difference detectable?

The \(penetration\) \(distance\) \(\eta\) in a finite potential well is the distance at which the wave function has decreased to 1/\(e\) of the wave function at the classical turning point: $$\psi(x = L + \eta) = {1\over e} \psi(L)$$ The penetration distance can be shown to be $$\eta = {\hslash \over \sqrt{2m(U_0 - E)}}$$ The probability of finding the particle beyond the penetration distance is nearly zero. (a) Find \(\eta\) for an electron having a kinetic energy of 13 eV in a potential well with \(U_0\) = 20 eV. (b) Find \(\eta\) for a 20.0-MeV proton trapped in a 30.0-MeV-deep potential well.

The ground-state energy of a harmonic oscillator is 5.60 eV. If the oscillator undergoes a transition from its \(n\) = 3 to \(n\) = 2 level by emitting a photon, what is the wavelength of the photon?

Protons, neutrons, and many other particles are made of more fundamental particles called \(quarks\) and \(antiquarks\) (the antimatter equivalent of quarks). A quark and an antiquark can form a bound state with a variety of different energy levels, each of which corresponds to a different particle observed in the laboratory. As an example, the \(\psi\) particle is a low-energy bound state of a so-called charm quark and its antiquark, with a rest energy of 3097 MeV; the \(\psi\)(2S) particle is an excited state of this same quark- antiquark combination, with a rest energy of 3686 MeV. A simplified representation of the potential energy of interaction between a quark and an antiquark is \(U(x) = A\mid x \mid\) , where \(A\) is a positive constant and \(x\) represents the distance between the quark and the antiquark. You can use the WKB approximation (see Challenge Problem 40.64) to determine the bound-state energy levels for this potential-energy function. In the WKB approximation, the energy levels are the solutions to the equation $$\int ^b _a \sqrt{2m[E - U(x)]} dx = {nh \over 2} \space (n = 1, 2, 3, . . .)$$ Here \(E\) is the energy, \(U(x)\) is the potential-energy function, and \(x\) = a and \(x\) = b are the classical turning points (the points at which \(E\) is equal to the potential energy, so the Newtonian kinetic energy would be zero). (a) Determine the classical turning points for the potential \(U(x) = A \mid x \mid\) and for an energy \(E\). (b) Carry out the above integral and show that the allowed energy levels in the WKB approximation are given by $$E_n = {1 \over2m} ( {3mAh \over 4} ) ^{2/3} n^{2/3} \space (n = 1, 2, 3, . . .)$$ (\(Hint\): The integrand is even, so the integral from \(-x\) to \(x\) is equal to twice the integral from 0 to \(x\).) (c) Does the difference in energy between successive levels increase, decrease, or remain the same as \(n\) increases? How does this compare to the behavior of the energy levels for the harmonic oscillator? For the particle in a box? Can you suggest a simple rule that relates the difference in energy between successive levels to the shape of the potential-energy function?

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