# Homework Help: Tunneling with an alpha particle

1. Apr 10, 2015

### Differentiate1

1. The problem statement, all variables and given/known data

In a simple model for a radioactive nucleus, an alpha particle (m = 6.64×10^−27kg) is trapped by a square barrier that has width 2.0*10^-15 meter and height 30.0 MeV.

What is the tunneling probability if the energy of the alpha particle is 18.0MeV below the top of the barrier?

E = 18 MeV
U = 30 MeV

m = 6.64*10^-27 kg
L = 2.0*10^-15 m

ћ = 1.055*10^-34 Js

2. Relevant equations

Probability of Tunneling

3. The attempt at a solution

G = 16(18/30)(1-(18/30)) = 3.84

U - E = 12*10^6 eV = 1.92*10^-12 J

κ = sqrt(2 * 6.64*10^-27 * 1.92*10^-12) / 1.055*10^-34
= 1.51*10^15 m^-1

L = 2*10^-15 m

---------------------------------------------------------------------------------------

e^(-2κL) = e^(-2 * 1.51*10^15 * 2*10^-15) = .0023

T = G * .0023 = 3.84 * .0023
= 9.0*10^-3

I've tried solving this problem numerous times and always end up with the same value listed above. Any observation on what went wrong would be appreciated. Thanks in advance.

2. Apr 11, 2015

### Orodruin

Staff Emeritus
E is defined as being 18 MeV below the barrier, not as being 18 MeV. This does not matter for G but it does for kappa.

3. Apr 11, 2015

### Differentiate1

Would that simply mean for kappa, instead of U - E, since it's below the barrier, it would be E - U?
My book defines U - E as being the additional KE needed to climb over the barrier.

Actually, that won't work algebraically since the numerator will be the square root of a negative value.
I am uncertain about this--maybe if the particle tunnels below, it means U - (-E)?

Last edited: Apr 11, 2015
4. Apr 11, 2015

### Orodruin

Staff Emeritus
No, it means that U-E is 18 MeV.

5. Apr 11, 2015

### Differentiate1

Can you please explain the concept behind why that's the case?

Last edited: Apr 11, 2015
6. Apr 11, 2015

### Orodruin

Staff Emeritus
Because this is what the problem states:

7. Apr 11, 2015

### Differentiate1

Thank you for your assistance!