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.

 

[member 553083]

Kinetic energy is the energy that an object has due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its current velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest.For a proton, the kinetic energy K (sub t) it will have on the other side of the barrier if it tunnels through the barrier can be calculated using the following equation:K (sub t) = (1/2)mv^2where m is the mass of the proton and v is the velocity of the proton.Similarly, the kinetic energy K (sub r) it will have if it reflects from the barrier can be calculated using the same equation.