Revisiting Basic Nuclear Physics (3): Nuclear Fission

 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 consider first 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:

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) –  (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:

 

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

 

Where:

 

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 decays:

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

 

 ₄Be ⁷ (_-1 e⁰  , u) ₃ Li ⁷    

 

But 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 = [M_a + M_X  - M_Y – M_b]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 = [M_a + M_X  - M_Y – M_b]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

 

Example Problem 2:

 

We next use the example of the fission of U 235 illustrated in the top graphic 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:

Solution:

   ₀ n ¹     = 1.009u

 

₉₂U ²³⁵   = 235.123u

 

₄₂ Mo ⁹⁵   =  94.945u

 

₅₇ La ¹³⁹   =  138.955u

 

The left side of the reaction (sum of reactants) yields:

 

235.123u

+

1.009u

______

236.132u

 

The right side of the reaction yields:

 

95.945u

+

138.955u

+

2.018u

______

235.918u

 

Then:  Q = [(M + m) – (M’ + M” + m’) ]c²

 

Where:

 

(M + m) =  236.132u

 

And: (M’ + M” + m’) = 235.918u

 

Therefore:

 

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

 

Q = (0.214u) (931 MeV/u) =  198 MeV

 

A key point to note in this result is that Q > 0, indicating an exothermic reaction (energy given off) and that the kinetic energy of the particles is greater than the initial energy.

Suggested 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)  . 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 ⁶ (p, a) H ³

 

c) Li ⁷ (p, n) Be ⁷

d) F ¹⁹ (p, a) O ¹⁶

 Useful atomic masses:  C ¹³  (13.003355u), O ¹⁸  (17.999159u), N ¹⁴  (14.003074u), F ¹⁸ (18.00937u).