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Chapter 7: Problem 87

An electron with kinetic energy \(2.0 \mathrm{MeV}\) encounters a potential energy barrier of height \(16.0 \mathrm{MeV}\) and width 2.00 nm. What is the probability that the electron emerges on the other side of the barrier?

Short Answer

Expert verified
The probability for the electron to emerge on the other side of the barrier is given by T, which is calculated using the formula for quantum tunneling. The exact value can be obtained after substituting the known quantities into the formula.

Step by step solution

01

Identifying known quantities

From the problem, we identify the given quantities: kinetic energy of the electron (E) is \(2.0 \mathrm{MeV}\), potential energy barrier (V) is \(16.0 \mathrm{MeV}\), and width of the barrier (a) is \(2.00 \mathrm{nm}\).
02

Converting units

All quantities must be expressed in same unit system. Therefore, convert energy from MeV to J by using the conversion factor \(1.6 x 10^{-13} J/MeV\) and distance from nm to m by using the conversion factor \(10^{-9} m/nm\).
03

Calculate the probability

Use the formula for quantum tunneling to calculate the probability (T). For a barrier where \(V >> E\), the transmission coefficient (T) is given by: \(T = \exp(-2a \sqrt{2m(V - E)/(\hbar)}))\), where m is the mass of the electron (\(9.1 x 10^{-31} kg\)), \(\hbar\) is the reduced Planck’s constant (\(1.05 x 10^{-34} Js\)). Substitute the all known values into the formula to calculate T.
04

Interpret the result

The value of T represents the probability that the electron will overcome the barrier and emerge on the other side. If T is small, it means the probability is low.

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Most popular questions from this chapter

An electron confined to a box of width 0.15 nm by infinite potential energy barriers emits a photon when it makes a transition from the first excited state to the ground state. Find the wavelength of the emitted photon.

Vibrations of the hydrogen molecule \(\mathrm{H}_{2}\) can be modeled as a simple harmonic oscillator with the spring constant \(\quad k=1.13 \times 10^{3} \mathrm{N} / \mathrm{m} \quad\) and \(\quad\) mass \(m=1.67 \times 10^{-27} \mathrm{kg} .\) (a) What is the vibrational frequency of this molecule? (b) What are the energy and the wavelength of the emitted photon when the molecule makes transition between its third and second excited states?

. Estimate the ground state energy of the quantum harmonic oscillator by Heisenberg's uncertainty principle. Start by assuming that the product of the uncertainties \(\Delta x\) and \(\Delta p\) is at its minimum. Write \(\Delta p\) in terms of \(\Delta x\) and assume that for the ground state \(x \approx \Delta x\) and \(p \approx \Delta p\) then write the ground state energy in terms of \(x .\) Finally, find the value of \(x\) that minimizes the energy and find the minimum of the energy.

If a classical harmonic oscillator can be at rest, why can the quantum harmonic oscillator never be at rest? Does this violate Bohr's correspondence principle?

Is it possible to measure energy of \(0.75 \hbar \omega\) for a quantum harmonic oscillator? Why? Why not? Explain.
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