*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:

^{2}

**:**

*Decay and Nuclear Fission Reactions***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 for Polonium:

_{84 }

*Po*

^{210}

*®*

_{82}Pb^{206}+_{2}He^{4}
If
the mass of

**= 209.982u, the mass of**_{84 }Po^{210}**= 205.969u, and**Po _{82}^{206}**= 4.002u.**_{2}He^{4}
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) – (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: for Beryllium 7

_{4}Be^{7}+_{-1}e^{0}

*®*

_{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: c

^{2}= 931.5 MeV/u
Again,

**Q =**

**[(total rest mass before decay) –**

**(total rest mass after decay )] c**

^{2}**931.5 MeV/u = 1.996u**

*]*
Q
= (1.996u) 931.5 MeV/u = 1859 MeV

*:*

**Compressed notation for nuclear decays****(**

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

_{-1}e^{0}

*u*

*)*_{3 }Li^{7}*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).

a
+ X ® Y + b

X
(a,b) Y

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 }=[

**c***7.016004 u + 1.007825 u - 4.002603u - 4.002603u]*^{2}

*= [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)

Po _{84 }^{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 energy*”**associated with the neutron. Estimate the value of this energy if the atomic mass of***S*_{n}**= 42.98780u, and***Ca*^{43}**= 41.958625u, and take the neutron mass = 1.008665u.***Ca*^{42}
4.
For each of the following reactions, write out the full nuclear equation and
find the Q of the reaction:

a)

**(Note: d is for***H*^{2}(d, p) H^{3 }*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:

_{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:

^{ }

_{0 }n^{1}= 1.009u

_{92}

*U*^{235}= 235.123u

_{42}

*Mo*^{95}= 94.945u

_{57}

*La*^{139}= 138.955u^{}

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