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Ch 40: Quantum Mechanics I: Wave Functions
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 40, Problem 29

An electron with initial kinetic energy 6.06.0 eV encounters a barrier with height 11.011.0 eV. What is the probability of tunneling if the width of the barrier is (a) 0.800.80 nm and (b) 0.40 0.40 nm?

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Step 1: Recognize that this is a quantum tunneling problem, where the probability of tunneling through a potential barrier is given by the formula: \( T = e^{-2 \kappa L} \), where \( \kappa = \sqrt{\frac{2m(U - E)}{\hbar^2}} \). Here, \( m \) is the mass of the electron, \( U \) is the barrier height, \( E \) is the electron's energy, \( \hbar \) is the reduced Planck's constant, and \( L \) is the width of the barrier.
Step 2: Convert the given energies from electron volts (eV) to joules (J) using the conversion factor \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). Calculate \( U - E \) in joules, where \( U = 11.0 \text{ eV} \) and \( E = 6.0 \text{ eV} \).
Step 3: Calculate \( \kappa \) using the formula \( \kappa = \sqrt{\frac{2m(U - E)}{\hbar^2}} \). Use the mass of the electron \( m = 9.11 \times 10^{-31} \text{ kg} \) and \( \hbar = 1.055 \times 10^{-34} \text{ J·s} \). Substitute the value of \( U - E \) from Step 2 into this equation.
Step 4: For part (a), substitute \( L = 0.80 \text{ nm} = 0.80 \times 10^{-9} \text{ m} \) into the tunneling probability formula \( T = e^{-2 \kappa L} \). For part (b), repeat the calculation with \( L = 0.40 \text{ nm} = 0.40 \times 10^{-9} \text{ m} \).
Step 5: Simplify the expressions for \( T \) for both cases (a) and (b) to express the tunneling probabilities in terms of \( \kappa \) and \( L \). This will give the final expressions for the tunneling probabilities without numerical evaluation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Quantum Tunneling

Quantum tunneling is a phenomenon in quantum mechanics where a particle can pass through a potential energy barrier, even if its energy is less than the height of the barrier. This occurs due to the wave-like nature of particles, allowing for a non-zero probability of finding the particle on the other side of the barrier. The probability of tunneling decreases exponentially with increasing barrier width and height.

Kinetic Energy and Potential Energy

Kinetic energy is the energy possessed by an object due to its motion, while potential energy is the stored energy based on an object's position in a force field, such as gravitational or electric fields. In this context, the electron's initial kinetic energy (6.0 eV) is compared to the potential energy barrier (11.0 eV) it encounters, determining the likelihood of tunneling through the barrier.
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Barrier Width and Height in Tunneling

The width and height of a potential barrier significantly influence the tunneling probability of a particle. A wider barrier or a higher barrier reduces the probability of tunneling, as the particle's wave function decays exponentially within the barrier. The relationship can be quantitatively described using the Schrödinger equation, which provides a mathematical framework for calculating tunneling probabilities based on these parameters.
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Related Practice
Textbook Question

(a) An electron with initial kinetic energy 3232 eV encoun­ters a square barrier with height 4141 eV and width 0.250.25 nm. What is the probability that the electron will tunnel through the barrier?

(b) A proton with the same kinetic energy encounters the same barrier. What is the probability that the proton will tunnel through the barrier?

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Textbook Question

For the ground level of a harmonic oscillator, xpx=ħ/2∆x∆p_x = ħ/2. Do a similar analysis for an excited level that has quantum number nn. How does the uncer­tainty product xpx∆x∆p_x depend on nn?

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Textbook Question

(a) Find the excitation energy from the ground level to the third excited level for an electron confined to a box of width 0.3600.360 nm.

(b) The electron makes a transition from the n=1n = 1 to n=4n = 4 level by absorbing a photon. Calculate the wave­length of this photon.

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Textbook Question

An electron is in a box of width 3.0×10103.0\(\times\)10^{-10} m. What are the de Broglie wavelength and the magnitude of the momentum of the electron if it is in (a) the n=1n = 1 level; (b) the n=2n = 2 level; (c) the n=3n = 3 level? In each case how does the wavelength compare to the width of the box?

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Textbook Question

An electron with initial kinetic energy 5.05.0 eV encoun­ters a barrier with height U0U_0 and width 0.600.60 nm. What is the transmission coefficient if (a) U0=7.0U_0 = 7.0 eV; (b) U0=9.0U_0 = 9.0 eV; (c) U0=13.0U_0 = 13.0 eV?

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Textbook Question

While undergoing a transition from the n=1n = 1 to the n=2n = 2 energy level, a harmonic oscillator absorbs a photon of wavelength 6.506.50 μμm. What is the wavelength of the absorbed photon when this oscillator undergoes a transition (a) from the n=2n = 2 to the n=3n = 3 energy level and (b) from the n=1n = 1 to the n=3n =3 energy level?

(c) What is the value of (k/m)\(\sqrt{(k^{\prime}\)/m)}, the angular oscillation frequency of the corresponding Newtonian oscillator?

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