3. Nuclear
Reactions
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,
(total rest mass after
decay )] c2
Q
= (1.996u) 931.5 MeV/u = 1859 MeV
Compressed
notation for nuclear decays:
4Be 7 (-1 e0 , u) 3 Li 7
a
+ X ® Y + b
X
(a,b) Y
Q
= [Ma + MX - MY
– Mb]c2
3 Li 7 (p, a) 2 He 4
And
obtain the Q-value.
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
[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 energy” S 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
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