## Monday, October 20, 2014

### Introduction to Nuclear Physics (2)

Continued from first instalment:

3. Nuclear Reactions

The energy liberated in nuclear reactions is referred to the Q- value or the “Q of the reaction”. This is the total energy released in the reaction and is usually expressed as:

Q = [(M + m) – M’ – m’]c2

Where M + m denotes the sum of masses of the reactants and M’, m’  denotes the masses of the products. Q is obtained by using the atomic masses revealed when the reaction has been fully written out.   Before doing this it’s important to recognize the various kinds of reactions and then how Q is applied, including how to obtain the kinetic energy and the initial energy:

Decay and Nuclear Fission Reactions:

We first consider natural decay and also artificial nuclear fission reactions, e.g. produced by bombardment of a nucleus by a smaller one. There are basically two types of decay processes we will look at:

Alpha decay:  Z X A ®  Z-2 X A-4 + 2 He 4

Beta Decay: Z X A + -1 e0 ®  Z-1 X A   + u

Example: Find the Q-Value of the alpha decay for Polonium:

84 Po 210     ®  82 Pb 206     +    2 He 4

If the mass of  84 Po 210   = 209.982u,  the mass of  82 Po 206   = 205.969u, and  2 He 4 = 4.002u.

We confirm that typical for a-decay, the atomic weight A decreases by 4, and the atomic number Z by 2. Then we may write for the Q of the reaction:

Q = [(total rest mass before decay) –

(total rest mass after decay )] c2

Q = [(209.982u) –  (205.969u + 4.002u)]c2

Q = [(209.982u) –  (205.969u + 4.002u)]c2

Q = [(209.982u) –  (209. 971u)] 931 MeV/u

Q = [ 0.011u] 931 MeV/u = 10.24 MeV

We next seek to find the Q-value associated with the beta decay: for Beryllium 7

4 Be 7    +    -1 e0 ®   3 Li 7     +  u

Where the respective atomic masses are:

4 Be 7    = 9.012182 u

3  Li  7   = 7.016004 u

And we use: c2 = 931.5 MeV/u

Again,

Q =   [(total rest mass before decay) –

(total rest mass after decay )] c2

Q = [(9.012182 u - 7.016004] 931.5 MeV/u = 1.996u

Q =  (1.996u) 931.5 MeV/u = 1859 MeV

Compressed notation for nuclear decays:

An abbreviated, concise notation is often used to represent nuclear reactions.  Consider the case of the beta decay just analyzed. We may write this in concise form:

4Be 7 (-1 e0  , u) 3 Li 7

Note that this is more often employed for the bombardment of a particle than for simple decays. For example, in the illustration above the Beryllium isn’t being bombarded by anything – rather it is emitting an electron! (Though we can still use the concise notation to represent this so long as we understand the electron is a decay particle).

In general, for bombardment, given a target nucleus X bombarded by a particle a, yielding a nucleus Y and another particle b, e.g.

a + X ®  Y + b

We have in more concise form:

X (a,b) Y

The Q-value of the reaction can then be worked out on the basis of:

Q = [Ma + MX  - MY – Mb]c2

Example Problem:  Write out the nuclear reaction for:

3 Li 7 (p, a) 2 He 4

And obtain the Q-value.

Solution:

We know the p denotes the proton of hydrogen nucleus and a  is an alpha particle given off. We list the nuclear masses as follows:

3 Li  7   = 7.016004 u

p =  1H 1 = 1.007825 u

a  =  2 He 4  =  4.002603u

Then:

Q = [Ma + MX  - MY – Mb]c2   =

[7.016004 u + 1.007825 u - 4.002603u - 4.002603u] c2

= [8.023829 u -  8.005206 u] 931.5 MeV/u

So: Q = (0.018623 u) 931.5 MeV/u =17.3 MeV

Problems:

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.

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

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.

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 3    (Note: d is for deuterium or 1 H 2 )

b) Li 7 (p, n) Be 7

5. Refer to the example of the fission of U 235 illustrated (Fig. 1)  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