*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’]c^{2}

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} e^{0} *®** _{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 )] c ^{2}**

Q
= [(209.982u) – (205.969u + 4.002u)]c^{2}

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}
e^{0} *

*®*

_{3 }Li^{7}+

*u*Where:

_{4}* Be ^{7} =
9.012182 u*

_{3
}* Li ^{7 }= 7.016004 u*

And
we use: c^{2} = 931.5 MeV/u

Again,

**Q
= [(total rest mass before decay) –**

**
(total rest mass after decay )] c ^{2}**

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:

** _{4}Be ^{7}** (

**,**

_{-1}e^{0}

*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}]c^{2}

* 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

**is an alpha particle given off. We list the nuclear masses as follows:**

*a*_{3
}*Li
^{7
}= 7.016004 u*

p
= * _{1}H ^{1}*

*= 1.007825 u**a** = _{2} He ^{4 } = 4.002603u*

Then: Q
= [M_{a }+ M_{X} - M_{Y}
– M_{b}]c^{2}

=
[** 7.016004 u + 1.007825 u - 4.002603u - 4.002603u] **c

^{2}

*=
[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:

_{92}*U ^{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*

_{92}*U ^{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’) ]c^{2}

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

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),

**(17.999159u),**

*O*^{18}**(14.003074u),**

*N*^{14}**(18.00937u).**

*F*^{18}
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