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A particle with energy \(E=5 \mathrm{eV}\) approaches a barrier of height \(U=8 \mathrm{eV}\) Quantum mechanically there is a nonzero probability that the particle will tunnel through the barrier. If the barrier height is slowly decreased, the probability that the particle will deflect off the barrier will a) decrease. b) increase. c) not change.

Short Answer

Expert verified
Answer: The probability of a particle deflecting off a barrier decreases as the barrier height decreases due to the increased likelihood of quantum tunneling.

Step by step solution

01

Understand Quantum Tunneling

Quantum tunneling is a quantum mechanical phenomenon in which a particle can pass through a potential barrier even if it has less energy than the barrier height. This occurs because of the wave nature of particles in quantum mechanics and their associated probability distributions.
02

Analyze the given situation

In this problem, we have a particle of energy E = 5 eV approaching a barrier of height U = 8 eV. The fact that the particle's energy is smaller than the barrier height signifies that it could be possible for the particle to tunnel through the barrier, due to the quantum mechanical nature of particles.
03

Determine the effect of decreasing barrier height on tunneling probability

When the barrier height is decreased, the probability of the particle tunneling through the barrier increases. This is because the lower barrier height allows for a greater transmission of the particle's probability wave through the barrier. This increased transmission results in an increased likelihood of the particle being found on the other side of the barrier.
04

Conclude the effect on probability of the particle deflecting off the barrier

Since the probability of the particle tunneling through the barrier has increased as the barrier height decreases, the probability of the particle deflecting off the barrier will conversely decrease. Thus, the correct answer is: a) decrease.

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Most popular questions from this chapter

A particle of mass \(m\) is in the potential \(U(x)=U_{0} \cosh (x / a),\) where \(U_{0}\) and \(a\) are constants. Show that the ground-state energy of the particle can be estimated as $$E_{0} \cong U_{0}+\frac{1}{2} \hbar\left(\frac{U_{0}}{m a^{2}}\right)^{1 / 2}$$

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