Barrier Tunneling and Kinetic Energy

[nickm4]

1. Homework Statement

b). Find the kinetic energy K (sub t), the proton will have on the other side of the barrier if it tunnels through the barrier.

c) Find the kinetic energy K (sub r), it will have if it reflects from the barrier.

Variables:

Transmission Coefficient (T)

T= e^-2bL
T was found to be T= e^-11.617 or (9.011*10^-6)
e= 2.718...
L= length of the barrier which is given as 10fm or (10.0*10^-15m)

b= sqrt(((8pie^2)(m)(U(sub b)-E))/(h^2))

m= mass of proton(1.673*10^-27kg)
Ub= height of the potential barrier(given= 10MeV)
E= energy of the proton (given= 3MeV)
h= plank's constant (6.62*10^-34)

2.

Homework Equations

T= e^-2bL
b= sqrt(((8pie^2)(m)(U(sub b)-E))/(h^2))

The Attempt at a Solution

I solved the first part of the question to find the transmission coefficient, T. But I'm not sure how Kinetic energy is related. Other than through b.

This question is taken from " Fundementals of Physics" Halliday/Resnick 7th ED. Question: 38-63

Thanks Tons.

 

[Marthius]

In both cases (reflection and tunneling) the energy remains unchanged. By conservation of energy the energy cannot change because it is not going anywhere. This holds for electrons and other particles as well.

 

[Moetasim]

I would like to provide a response to the content regarding barrier tunneling and kinetic energy. First, it is important to understand the concept of barrier tunneling. When a particle, in this case a proton, encounters a potential barrier, it has a certain probability of passing through the barrier rather than being reflected back. This phenomenon is known as barrier tunneling.

In this specific question, we are given the transmission coefficient (T) which represents the probability of the proton tunneling through the barrier. Using this information, we can calculate the kinetic energy (K) of the proton on the other side of the barrier using the equation K = (1-T)E, where E is the initial energy of the proton. This means that if the proton successfully tunnels through the barrier, it will have a kinetic energy equal to the initial energy minus the energy reflected back.

On the other hand, if the proton is reflected back from the barrier, it will have a kinetic energy equal to the initial energy (E). This is because the transmission coefficient is equal to zero and therefore there is no energy loss due to tunneling.

To further understand how kinetic energy is related to barrier tunneling, it is important to look at the equations provided in the problem. The parameter b, which is used to calculate the transmission coefficient, is also used in the equation for kinetic energy as b = sqrt(((8pie^2)(m)(U(sub b)-E))/(h^2)). This shows that as b increases, the transmission coefficient decreases, meaning that the probability of tunneling decreases. This results in a higher kinetic energy for the proton if it is reflected back from the barrier.

In conclusion, barrier tunneling and kinetic energy are closely related in this scenario. The transmission coefficient determines the probability of tunneling and subsequently, the kinetic energy of the proton on the other side of the barrier. I hope this explanation helps to clarify any confusion and provides a better understanding of these concepts.