A 3.0 MeV proton is incident on a potential energy barrier...

A $3.0 \mathrm{MeV}$ proton is incident on a potential energy barrier of thickness $10 \mathrm{fm}$ and height $10 \mathrm{MeV}$. What are (a) the transmission coefficient $T,$ (b) the kinetic energy $K,$ the proton will have on the other side of the barrier if it tunnels through the barrier, and (c) the kinetic energy $K$, it will have if it reflects from the barrier? A $3.0 \mathrm{MeV}$ deuteron (the same charge but twice the mass as a proton) is incident on the same barrier. What are (d) $T,(\mathrm{e}) K_{\mathrm{N}}$ and (f) $K_{r} ?$

Key Concepts

Barrier Thickness and Height Effects

Both the thickness and height of a potential barrier greatly influence tunneling probability. A thicker barrier or a higher barrier relative to the incoming particle’s energy leads to a more pronounced exponential decrease in the wavefunction within the barrier, thereby reducing the transmission coefficient. These parameters are key in determining how likely it is for a particle to tunnel through the barrier.

Dependence on Particle Mass

The mass of the particle plays an important role in tunneling phenomena. Heavier particles have wavefunctions that decay more rapidly inside the barrier, which reduces the tunneling probability. This mass dependence is crucial when comparing tunneling behaviors of particles with different masses even if they experience the same potential barrier.

Conservation of Energy

The principle of conservation of energy holds in quantum tunneling, meaning that the total energy of the particle is conserved throughout the process. For a tunneling particle, if the barrier is static and the surrounding potential remains the same, the kinetic energy of the particle after tunneling is effectively restored to its original energy, while in the case of reflection, the kinetic energy remains unchanged.

Transmission Coefficient

The transmission coefficient is a measure of the probability that a particle will tunnel through a potential energy barrier. It depends on factors such as the particle’s energy, the barrier’s height and thickness, and the particle’s mass. In quantum mechanical problems, the coefficient is calculated from the exponential decay of the wavefunction in the barrier region and plays a central role in quantifying tunneling rates.

Quantum Tunneling

Quantum tunneling is the phenomenon in which a particle can cross a potential energy barrier even when its energy is lower than the height of the barrier. This occurs because the wave-like nature of particles in quantum mechanics allows for a finite probability of the particle being found on the other side of a classically forbidden region, as described by the solutions to the Schrödinger equation.

Potential Energy Barrier

A potential energy barrier is a region in space where the potential energy is higher than the energy of an incident particle, creating a classically forbidden zone. In quantum mechanics, even when the particle’s energy is insufficient to overcome the barrier classically, it may still have a finite chance of appearing on the other side via tunneling.

Transcript

00:01 For this problem on the topic of photons, we have a 3 mega electron -volt proton incident on a potential energy barrier, which is a thickness of 10 femtometers and a height of 10 mega -electron volts.

00:11 We want to find the transmission coefficient t, the kinetic energy k that the proton will have on the other side of the barrier if it tunnels through the barrier.

00:19 The kinetic energy it will have if it reflects from the barrier.

00:22 And then we have a 3 -mega electron volt deuteron, which has the same mass, same charge, but twice the mass of a proton incident on the same barrier.

00:30 We again want to find the transmission coefficient and the kinetic energy each time.

00:39 So for part a, we know the transmission coefficient t for a particle of mass m and energy e that is incident on a barrier of height ub, and with l is given by the equation t is equal to e to the minus 2b times l, where b is equal to the square root of 8 pi 3 .3.

01:04 Squared m into u b minus e divided by h squared and so for the proton we have this value to be the square root of 8 pi squared times the mass of a proton 1 .6726 times 10 to the minus 27 kg times 10 mega electron volts minus 3 mega electron volts multiplied by 1 .6022 times 10 to the minus 13 joules per mega electron volt and all of this divided by planks constant 6 .6216 .61 times 10 to the minus 34 jule seconds squared.

02:30 And so we get b to be 5 .8082 times 10 to the power 14 per meter, which gives us b times l to be 5 .8082 times 10 to the 14 per meter times 10.

02:58 Times 10 to the minus 15 meters, which is 5 .8082, which means that the transmission coefficient t is e to the minus 2 times 5 .8082, which gives the transmission coefficient of 9 .02 times 10 to the minus 6.

03:38 Now to find the kinetic energy, we know the mechanical energy is conserved.

03:44 So before the proton reaches the barrier, it has a kinetic energy of 3 mega electron volts and a potential energy of 0.

03:50 After passing through the barrier, the proton again has a potential energy of 0.

03:55 And so the kinetic energy k is, again, 3 mega electron volts...

Please give Ace some feedback