## Friday, February 4, 2022

### Solutions To Nuclear Fission Problems (Nuclear Physics, Part 3)

1.Write out the full nuclear reaction for:

13 Al 27 (a, n) 15 P 30

Thence, or otherwise, find the Q of the reaction.

Solution:

We write out the reaction from its condensed form given:

13 Al 27  + 2 He 4      ®  15 P 30 0 n 1

Then: The Q of the reaction is defined according to:

Q = [Ma + MX  - MY – Mb]c2

Where:

0 n 1 = Mb     = 1.008665u

a = MX  =  2 He 4  =  4.002603u

15 P 30    = MY  =   29.9783138u

13 Al 27  = M 26.9815386u

Q =   [26.9815386u  +   4.002603u  - 29.978313 u  - 1.008665u] 931 MeV/u

=   [30.984141 u  -   30.986978]u (931 MeV/u)=   - 2.64 MeV

(Negative sign denotes an endothermic reaction)

2. Identify the missing ‘X’ in each of the following:

a) 84 Po 215   ®  a

b)   14 (a, X)  O 17

c) 48 Cd 109    +   ®   47  Ag 109     +  u

Solutions:

T   a) This is an example of alpha decay defined by:

Z X A ®  Z-2 X A-4 + 2 He 4

Then:  A – 4 = 215 – 4 = 211

And:  Z – 2 =  84 – 2 = 82

So: X =   Pb (lead) e.g.  82 Pb 211

b) This is a condensed form for the reaction which would be written out:

7 N 14   + a ®  X + 8 O 17

Or:  7 N 14    + 2 He 4     ® Z1 +Z2  X A1 + A2  + 8 O 17

Now: A1 + A2 =  14 + 4 = 18 and Z1 + Z2 = 7 + 2 = 9

So:

Z1 +Z2  X A1 + A2  =   9  18

Which is fluorine,  or X = F (e.g. 9 F 18 )

c)   c) By inspection we see the mass number Z remains unchanged but the atomic number decreases by one. This must be an example of beta decay, defined:

Z X A + -1 e0 ®  Z-1 X A   + u

Hence, X = -10

3)  . For each of the following reactions, write out the full nuclear equation and find the Q of the reaction:

a) 2 (d, p)  H 3    (Note: d is for deuterium or 2 )

b) Li 6 (p, a) H 3

c) Li 7 (p, n) Be 7

d) 19 (p, a) O 16

Solutions:

a)  1H 2  +   1H 2  ®  1H +   1H 3

Q = [Ma + MX  - MY – Mb]c2

Then:

1 H 2 = Ma     = 2.01410 u

1 H 2  = MX  =  2.01410 u

1H 1 = M = 1.007825 u

1 H 2  = MY  =   3.01604 u

Therefore:

Q = [2(2.01410 u)  - 1.007825 u - 3.01604 u] 931 Mev/u =

[(4.0282) – 4.0238)] 931 Mev/u  =  0.0044 (931 Mev/u )

Q  = 4.09 MeV

Solutions to (b) - (d) follow analogous approaches,  simply substituting for the respective  Ma ,  MX , MY, Mb  in the specific nuclear reactions.  E.g. for (c):

Li 7 (p, n) Be 7

For:   3 Li 7   +   1H 1  ®    0 n 1  +   4 Be 7

Q = [M+ MX  - MY – Mb]c2

Ma  =  7.016004 u  (for  Li  7  )

MX  = 1.007825 u  (for  11 )

MY =  1.009u    (for   0 n 1  )

Mb  =  7.0169287 u  (for   4 Be )

Q = [7.016004 u 1.007825 u  - 1.009u – 7.0169287 u] c2

Q= - 0.004u (931 Mev/u )  =  -3.72 MeV