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.
Solution: We write out the
reaction from its condensed form given:
13
Al
27 + 2 He 4 ® 15 P 30 + 0 n 1
Then:
The Q of the reaction is defined according to:
Q
= [Ma + MX - MY
– Mb]c2
Where:
0
n
1 = Mb
= 1.008665u
a = MX = 2
He 4 = 4.002603u
15 P 30 =
MY = 29.9783138u
(from Wikipedia table, Googled)
13
Al
27 = Ma = 26.9815386u (from
Wikipedia table – Googled)
Q = [26.9815386u + 4.002603u
- 29.978313
u - 1.008665u] 931 MeV/u
= [30.984141 u
- 30.986978]u (931 MeV/u)= - 2.64 MeV
(Negative
sign denotes an endothermic 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
Solutions:
a)
This
is an example of alpha decay defined by:
Z
X
A ® Z-2 X A-4 + 2
He 4
Then: A – 4 = 215 – 4 = 211
And: Z – 2 =
84 – 2 = 82
So:
X = Pb (lead) e.g. 82 Pb 211
b)
This is a condensed form for the reaction which would be written out:
7
N
14 + a ® X + 8 O 17
Or: 7 N 14 + 2 He 4 ® Z1 +Z2 X A1 + A2 + 8 O 17
Now: A1 + A2 = 14 + 4 = 18 and Z1 + Z2 = 7 +
2 = 9
So:
Z1 +Z2 X A1 + A2 = 9
X18 which is fluorine, or X = F (e.g. 9
F 18 )
b)
By
inspection we see the mass number Z remains unchanged but the atomic number
decreases by one. This must be an example of beta decay, defined:
Z X A + -1
e0 ® Z-1 X A + u
Hence,
X = -1 e0
So
that: 48 Cd 109
+-1 e0 ® 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.
S
n
is just another variant of the Q for a
reaction where:
S
n
=
[(total rest mass before decay)
–
(total rest
mass after decay )] c2
S
n
=
[(42.98780 u) - (41.958625u + 1.008665u)] 931 Mev/ u =
[(42.98780 u) - (41.958625u + 1.008665u)] 931 Mev/ u =
(42.98780
u – 42.96729u ) 931 Mev/ u =
0.0205 u (931 Mev/ u ) = 19.09 MeV
0.0205 u (931 Mev/ u ) = 19.09 MeV
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
Solution:
a) 1H 2 + 1H
2 ® 1H 1 +
1H 3
Q
= [Ma + MX - MY
– Mb]c2
1
H
2 = Ma
= 2.01410 u
1
H
2 = MX = 2.01410
u
1H 1 = Mb = 1.007825 u
1
H 2 = MY = 3.01604
u (from Wikipedia)
Therefore:
Q
= [2(2.01410
u) - 1.007825 u - 3.01604 u] 931 Mev/u
= [(4.0282) – 4.0238)] 931 Mev/u =
0.0044 (931 Mev/u ) = 4.09 MeV
5.
Refer to the example of the fission of U 235 illustrated 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
Q
= [Ma + MX - MY
– Mb]c2
Where:
92U 235 = Ma = 235.123u
0
n
1 = MX = 1.009u
42 Mo 95 + 57
La 139 = Mb =94.945u + 138.955u = 233.900 u
2 (0 n 1) = MY = 2(1.009u)
= 2. 018 u
Then:
Q
= [(235.123u + 1.009u) –
(233.900 u +2. 018 u)] c2
Q
= [(236.132)
- (235.918)] 931MeV/u =
0.214(931MeV/u)
Q = 199.2 MeV
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