## Thursday, October 23, 2014

### Solutions to Problems - Intro. To Nuclear Physics (2)

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  (from Wikipedia table, Googled)

13 Al 27  = M =  26.9815386u (from Wikipedia table – Googled)

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   ®  X + a

b)   N 14 (a, X)  O 17

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

Solutions:

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  X18  which is fluorine,  or X = F (e.g. 9  F 18 )

b)     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 = -1 e0

So that: 48 Cd 109    +-1 e0   ®   47  Ag 109     +  u

3. Consider a process of neutron removal whereby:

20 Ca 43    ®   20 Ca 42     + 0 n 1

We wish to find the “separation energyS n associated with the neutron.  Estimate the value of this energy if the atomic mass of Ca 43     = 42.98780u, and  Ca 42  = 41.958625u, and take the neutron mass = 1.008665u.

S n  is just another variant of the Q for a reaction where:

S n  =   [(total rest mass before decay) –
(total rest mass after decay )] c2

S n  =

[(42.98780 u) -  (41.958625u + 1.008665u)] 931 Mev/ u  =

(42.98780 u – 42.96729u ) 931 Mev/ u  =

0.0205 u (931 Mev/ u ) = 19.09 MeV

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

a) H 2 (d, p)  H   (Note: d is for deuterium or 1 H 2 )

b) Li 7 (p, n) Be 7

Solution:

a)  1H 2  +   1H 2  ®  1H +   1H 3

Q = [Ma + MX  - MY – Mb]c2

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 (from Wikipedia)

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  ) = 4.09 MeV

5. Refer to the example of the fission of U 235 illustrated at the outset of Part 1. This fission reaction may be written:

92U 235   +  0 n 1     ®  42 Mo 95     +    57 La 139 + 2 (0 n 1)

Find the Q-value of this reaction, given these atomic masses:

0 n 1     = 1.009u

92U 235   = 235.123u

42 Mo 95   =  94.945u

57 La 139   =  138.955u

Q = [Ma + MX  - MY – Mb]c2

Where:

92U 235    = Ma   = 235.123u

0 n 1       = MX  = 1.009u

42 Mo 95  +  57 La 139  = M =94.945u + 138.955u =  233.900 u

2 (0 n 1)  = MY  =   2(1.009u) =  2. 018 u

Then:

Q = [(235.123u + 1.009u) – (233.900 u +2. 018 u)] c2

Q =   [(236.132)  -  (235.918)] 931MeV/u  =  0.214(931MeV/u)

Q =  199.2 MeV