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Barrier Tunneling and Kinetic Energy

  • Thread starter nickm4
  • Start date
4
0


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.


2. Homework Equations

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

3. 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.
 

Answers and Replies

37
0
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.
 

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