Physics 3401, Chapter 7 Quiz

1. a. Calculate the ground state energy of a proton in an infinite-walled box in 3 dimensions
with side L= 10 fermi.

E1 for 3-D box = 3h^2/8mL^2=3(hc)^2/(8mc^2L^2)
E1=3*(1240 MeV fm)^2/(8*938 MeV*(10 fm)^2) =    6.147 MeV

b. Calculate the ground state energy of an electron in the same box.
E1=3*(1240 MeV fm)^2/(8*.511 MeV*(10 fm)^2) =    11283.8 MeV

2. For an electron contained in a one-dimensional box of dimension L=.3 nm:

a. Calculate the ground state energy if the walls are infinite.
E1=(hc)^2/(8*mc^2*L^2)=(1240 eV nm)^2/(8*.511 MeV*(.3 nm)^2)=
E1= 4.179 eV

b. Calculate the ground state energy if the walls are 100 eV high. This involves the calculation
of a new effective dimension L by calculating the distance at which the wavefunction drops to
1/e times its value inside the box.

Wavefuntion in barrier drops exponentially according to exp(-ax)
a=sqrt((2*mc^2/(hbar*c)^2)*(U-E1))=sqrt((2*.511MeV)/(197 eV nm)^2)*(100-4.179))
a= 50.233 1/nm
add to each side of box  the amount 1/a =    0.020 nm
New box width 0.340 nm
New ground state energy  = (1240 eV nm)^2/(8*.511 MeV*(.34 nm)^2)=
New E1 =  3.254 eV

3. The calculation of tunneling probability for an alpha particle out of a nucleus involves
approximating the coulomb barrier with a series of barriers of different height. An alpha
particle has energy 8 MeV inside the nucleus. Approximate the barrier with two
rectangular barriers of width 5 fm, the first at height 20 MeV and the second at height
10 MeV. Calculate the tunneling probabability for the alpha particle. The mass energy
of an alpha particle is 3727 MeV.

The wavefunction drops off exponentially in the barrier.
exp(-sqrt((2*mc^2)/(hbar*c)^2*(U-E))*x)

For the first barrier:
exp(-sqrt((2*3727 MeV)/(197 MeV fm)^2*(20-8)MeV)*x)=    5.05E-04
The probability is the square of the ratio of wavefunctions, so prob =    2.55E-07

For the second barrier:
exp(-sqrt((2*3727 MeV)/(197 MeV fm)^2*(10-8)MeV)*x)=    4.51E-02
The probability is the square of the ratio of wavefunctions, so prob =    2.03E-03

The product of the tunneling probabilities is    5.19E-10