## Tuesday, February 15, 2022

### Solutions To Nuclear Fusion Problems

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 macroscopic mass converted into energy and is a faithful reflection of all the fusion reactions underlying the  conversion.

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

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

d2 u I /dr2  +  a2  I = 0.

b) Show why the cosine solution needs to be discarded.

Solutions:

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:

I    =   B sin(ar)

(b) Answered at end of preceding, i.e. the cosine result is unphysical.