An electron with initial kinetic energy 6.0 6.0 eV encounters a barrier with height 11.0 11.0 eV. What is the probability of tunneling if the width of the barrier is (a) 0.80 0.80 nm and (b) 0.40 0.40 nm?

Hey everyone. So this problem is dealing with quantum mechanics and tunneling probabilities. Let's see what it's asking us. We have the kinetic energy of a proton that is 7.1 electron volts. The proton approaches a barrier with a potential energy of 12.2 electron volts. We're asked to determine the probability of tunneling for the following thicknesses of the barrier. Part one is two PICO meters and part two is four PICO meters. Our multiple choice answers are a 0.54 and 0.074. B. Both probabilities are zero. C both probabilities are one or D 0.54 and 0.23. The key to solving this problem is recall recalling our tunneling probability or T is given by the equation 16 E capital E for our um particle energy divided by you, not our potential energy multiplied by the quantity one minus E divided by U, not all of that multiplied by our lowercase E or natural exponent to the negative to KL where K is our wave number and L is our link. And so looking at the values that we have given to us in the problem we're given E, we're given U not and we're given L so we need to solve for K before we can solve for our tunneling probabilities. So we can recall that K, our wave number is given to us by the equation two multiplied by M are mass multiplied by U not minus E all divided by H bar squared where H bar is our reduced planks constant. Now, we're told that we have a proton. So we can recall that the mass of a proton is a constant and we can solve four K. So we have two, multiplied by that mass of a proton 1.67 times 10 to the negative 27 kg and then multiplied by U not our potential was given to us as 12.2 ev minus our proton energy 7.1 ev all divided by H bar, which is 1.055 times 10 to the negative 34. And we can recall that that is in units of dual seconds and that quantity is squared. And so the last thing for this step is to recognize that we have electron volts in our numerator joules in our denominator. And so we can use that conversion factor between electron volts and joules, which we, we recall is 1.6 times 10 to the negative 19 joules per electron volt. So plugging all of that into our calculators, we get a K of 4.95 times 10 to the 11. And that's in units of meters to the negative one. And now we have everything we need to solve for our tunneling probabilities. Recognizing that the only difference between parts one and parts two, parts one and two is this length, that length term. So we have tea or part one is equal to 16, multiplied by 7.1 ev divided by 12.2. EV multiplied by the quantity one minus E over UK not or E divided by U not. So that would be seven 0.1 ev divided by 12.2. EV multiplied by E to the negative two multiplied by K. So we just solve for that 4.95 times times the 11 meters the negative one. And then for part one, our length is two PICO meters or two times negative, sorry two times 10 to the negative 12 m. And so we plug that in and we get a T of two PICO meters of 0.54. That's our probability. And similarly for the tea tunneling probability in part two with four PICO meters, everything is the same except for that length of term. So we can copy everything the same. And then that length term is four PICO meters, four times 10 to the negative 12 m. I want to note that we didn't worry about changing our electron volts, converting our electron volts to Jews um for the tunneling probability function because the electron volts just cancel. So we don't need to worry about that. And when we plug that into our calculator, we get 0.074. And so those are the answers to this problem. And when we look at our multiple choice answers that aligns with answer choice A, so that's all we have for this one. We'll see you in the next video.

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Problem 29

Textbook Question

An electron with initial kinetic energy $6.0$ eV encounters a barrier with height $11.0$ eV. What is the probability of tunneling if the width of the barrier is (a) $0.80$ nm and (b) $0.40$ nm?

Verified step by step guidance

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.

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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(a) Find the excitation energy from the ground level to the third excited level for an electron confined to a box of width $0.360$ nm.

(b) The electron makes a transition from the $n equals 1$ to $n equals 4$ 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 $3.0 \times 10^{- 10}$ m. What are the de Broglie wavelength and the magnitude of the momentum of the electron if it is in (a) the $n equals 1$ level; (b) the $n equals 2$ level; (c) the $n equals 3$ level? In each case how does the wavelength compare to the width of the box?

Textbook Question

An electron is in a box of width 3.0*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 = 1 level; (b) the n = 2 level; (c) the n = 3 level? In each case how does the wavelength compare to the width of the box?

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Textbook Question

(a) An electron with initial kinetic energy $32$ eV encounters a square barrier with height $41$ eV and width $0.25$ 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?

Textbook Question

An electron with initial kinetic energy $5.0$ eV encounters a barrier with height $U sub 0$ and width $0.60$ nm. What is the transmission coefficient if (a) $U sub 0 equals 7.0$ eV; (b) $U sub 0 equals 9.0$ eV; (c) $U sub 0 equals 13.0$ eV?

Textbook Question

While undergoing a transition from the $n equals 1$ to the $n equals 2$ energy level, a harmonic oscillator absorbs a photon of wavelength $6.50$ $μ$ m. What is the wavelength of the absorbed photon when this oscillator undergoes a transition (a) from the $n equals 2$ to the $n equals 3$ energy level and (b) from the $n equals 1$ to the $n equals 3$ energy level?

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

Textbook Question

(II) What is the speed of an electron after being accelerated from rest by 28,000 V?

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Which of the following is a reasonable approximation of an inertial reference frame?

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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?

Key Concepts

Quantum Tunneling

Kinetic Energy and Potential Energy

Barrier Width and Height in Tunneling

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An electron with initial kinetic energy $6.0$ eV encounters a barrier with height $11.0$ eV. What is the probability of tunneling if the width of the barrier is (a) $0.80$ nm and (b) $0.40$ nm?