## Friday, October 31, 2014

### Selected Solutions to Nuclear Physics Problems (Part 3)

1) Calculate the wavelength of the gamma ray photon (in nm) which would be needed to balance the endothermic part of the triple –alpha fusion equation. (Recall here that 1 eV = 1.6 x 10-19 J)

Solution:

From the information given, we have:  E (g) =   hc/ l = 6.7 keV

So:

6.7 keV =  (6.62 x 10-34 J-s) (3 x 108 m/s ) /  l

Therefore:   l  =   (6.62 x 10-34 J-s) (3 x 108 m/s ) / 6.7 keV

Converting to consistent energy units, using 1 eV = 1.6 x 10-19 J:

l  =  (1.98 x 10-26 J-m)/  (1.07 x 10-16 J) =  1.85 x 10-10 m  =  185 nm

2) Verify the second part of the triple-alpha fusion reaction, especially the Q-value. Account for any differences in energy released by reference to the gamma ray photon coming off and specifically, give the wavelength of this photon required to validate the Q.

Solution: The 2nd part of the triple alpha fusion reaction is:

8Be + 4He  ®  12C + g  + 7.4 MeV

Q = [ (8.00531 u + 4.00260 u)– 12.0000 u] c2

Q =  [12.00795 u - 12.0000 u]   =  0.00795 u (931 MeV/u) = 7.4 MeV

The role (and value in energy) of the gamma ray photon can be obtained by using instead the value for carbon of 12.011 u and following the procedure shown in (1))

3) The luminosity or power of the Sun is measured to be L = 3.9 x 1026 watts.  Use this to estimate the mass (in kilograms) of the Sun that is converted into energy every second. State any assumptions made and reasoning.

Solution:

The luminosity is the same as the power or energy generated per unit time, thus:

L = E/ t  =  3.9 x 1026 J/s

The energy delivered per second then is:

3.9 x 1026 J   =   E  = m c2  so the mass converted to energy is:

m =   E/ c2     =  (3.9 x 1026 J)/ (3 x 108 m/s )2

= 4.3  x  109  kg

We have to assume the luminosity represents the actual macrosopic mass converted into energy and is a faithful reflection of all the fusion reactions underlying the  conversion.

4) In a diffusion cloud chamber experiment, it is found that alpha particles issuing from decay of U238 ionize the gas inside the chamber such that 5 x 10 3 ion pairs are produced per millimeter and on average each alpha particle traverses 25 mm. Estimate the energy associated with each detected vapor trail in the chamber  if each ion pair generates 5.2 x 10 -18   J:

Solution:

Let d be the  avg. length of each vapor trail, so :  d =   25 mm.

Each ion pair generates  5.2 x 10 -18 J, and  5 x 10 3 ion pairs are produced per millimeter, then:

E/ d  =  (5 x 10 3 ion pairs/mm) (5.2 x 10 -18 J)/ 25 mm = 1.04  x  - 15 J

5) When 118 Sn 50 is bombarded with a proton the main fission fragments are:

24 Na 11   and  94 Zr 40

The excitation energy necessary for passage over the potential barrier is:

e  >   3 ke2 Z2/ 5

Where the right hand side denotes the height of the Coulomb barrier.

a) What must this value be?
(Take   ke = 1.44 MeV/ fm)

Solution:
We have Z = 50 so that Z2  = 2500, then:

e   =    3 (1.44 MeV/ fm) (2500)/ 5  =   2160 MeV

b) Find the energy difference between the reactants and the products. (Take c2  = 931.5 MeV/ u)

Solution: The fission reaction can be written:

118 Sn 50      + 1H1 ®     24 Na 11   +  94 Zr 40

Therefore (Using atomic masses given in Wikipedia):

Q =

[(117.901606 u +   1.007825 u)  - 24.99096 u - 93. 907303 u ]  (931.5 MeV/u) =

[118.909431 -  118.898263] (931.5 MeV/u)  =

[0.011168] (931.5 MeV/u) = 10. 4 MeV

6)  a) Show that I    = A cos (ar)  + B sin(ar)

In  a solution of the reduced Schrodinger equation for the deuteron:

d2 u I /dr2  +  a2  I = 0.

Solution: We are given:

I    = A cos (ar)  + B sin(ar)

Then:  d/dr    = - a A sin (ar)  + a B cos(ar)

d2 u I /dr2  =   - a2 A cos(ar)  - a2 B sin (ar)   =

-  a2  [ A cos (ar)  + B sin(ar)]

Or:  d2 u I /dr2  =   -  a2  I

Transposing:

d2 u I /dr2  +  a2  I = 0.

Since R = u/ r  the cosine solution must be discarded, lest we get an unwanted infinity. This leaves:

u   =   B sin(ar)