Heavy nuclei often undergo alpha decay in which they emit an alpha particle (i.e., a helium nucleus). Alpha particles are so tightly bound together that it’s reasonable to think of an alpha particle as a single unit within the nucleus from which it is emitted. The probability that a nucleus will undergo alpha decay is proportional to the frequency with which the alpha particle reflects from the walls of the nucleus. What is that frequency (reflections/s) for a maximum-speed alpha particle within a 238U nucleus?
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35. Special Relativity
Inertial Reference Frames
Problem 9
Textbook Question
A finite potential well has depth U₀ = 2.00 eV. What is the penetration distance for an electron with energy (a) 0.50 eV, (b) 1.00 eV, and (c) 1.50 eV?

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Step 1: Understand the concept of penetration distance. In quantum mechanics, penetration distance refers to the distance a particle's wavefunction extends into a classically forbidden region (where the particle's energy is less than the potential energy). This is described by the exponential decay of the wavefunction in the forbidden region.
Step 2: Write the expression for the penetration distance. The penetration distance \( \lambda \) is given by \( \lambda = \frac{1}{\kappa} \), where \( \kappa \) is the decay constant. The decay constant \( \kappa \) is defined as \( \kappa = \sqrt{\frac{2m(U_0 - E)}{\hbar^2}} \), where \( m \) is the mass of the particle, \( U_0 \) is the potential well depth, \( E \) is the energy of the particle, and \( \hbar \) is the reduced Planck's constant.
Step 3: Substitute the known values into the formula for \( \kappa \). For an electron, the mass \( m \) is approximately \( 9.11 \times 10^{-31} \; \text{kg} \), and \( \hbar \) is approximately \( 1.055 \times 10^{-34} \; \text{J·s} \). Convert the given energies \( U_0 \) and \( E \) from electron volts (eV) to joules (J) using the conversion factor \( 1 \; \text{eV} = 1.602 \times 10^{-19} \; \text{J} \).
Step 4: Calculate \( \kappa \) for each case: (a) \( E = 0.50 \; \text{eV} \), (b) \( E = 1.00 \; \text{eV} \), and (c) \( E = 1.50 \; \text{eV} \). Use the formula \( \kappa = \sqrt{\frac{2m(U_0 - E)}{\hbar^2}} \) and substitute the corresponding values of \( U_0 \) and \( E \) for each case.
Step 5: Once \( \kappa \) is determined for each case, calculate the penetration distance \( \lambda \) using \( \lambda = \frac{1}{\kappa} \). This will give the penetration distance for the electron in each scenario.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Finite Potential Well
A finite potential well is a region in quantum mechanics where a particle experiences a potential energy lower than its surroundings. The well has a finite depth (U₀) and width, allowing for bound states where particles can exist with energies less than U₀. Outside the well, the potential energy is higher, leading to a phenomenon known as quantum tunneling, where particles can penetrate into classically forbidden regions.
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Quantum Tunneling
Quantum tunneling is a quantum mechanical phenomenon where a particle can pass through a potential barrier that it classically should not be able to surmount. This occurs due to the wave-like nature of particles, allowing them to have a non-zero probability of being found in regions where their energy is less than the potential energy. The penetration distance into the barrier is influenced by the energy of the particle and the height of the barrier.
Penetration Distance
The penetration distance refers to how far a particle can travel into a potential barrier before its probability of being found drops significantly. It is determined by the energy of the particle relative to the potential barrier height. For an electron in a finite potential well, the penetration distance increases as the energy of the electron approaches the depth of the well, allowing for a greater likelihood of finding the electron within the barrier.
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