3.016056 al
4. A tritium gas target (H}) is bombarded with a beam of protons (H}) of kinetic energy
3 MeV.
Determine Q value of the following reaction and specify the type of reaction.
H + {H → He + ón + Q
13
2016056 m
mini) = 1.008665 a.m.u.;
Answers
Answer:
A tritium gas target is bombarded with a beam of monoenergetic protons of kinetic energy K
1
=3MeV. The KE of the neutron emitted at 30
o
to the incident beam is K
2
? Find the value of
K
2
K
1
=
2
x
(approximately in whole number). Find x.
Atomic masses are H
1
=1.007276amu,n
1
=1.008665amu,
1
H
3
=3.016050amu,
2
He
3
=3.016030amu.
Hard
Solution
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Correct option is A)
Here the equation of nuclear reaction will be,
1
p
1
+
1
H
3
→
1
He
3
+
0
n
1
+Q
where Q=(m
p
+m
H
)−(m
He
+m
n
)=(1.007276+3.016050)−(3.016030+1.008665)
=−0.00137amu=−0.00137×931=−1.274MeV
The kinetic energy of emitted neutron is: K
2
=
u±(u
2
+v)
1/2
where u=
m
He
+m
n
m
p
m
n
K
1
cosθ=
3.016030+1.008665
1.007276×1.008665×3
cos30=0.375
and v=
m
He
+m
n
Qm
He
+K
1
(m
He
−m
P
)
=
3.016030+1.008665
(−1.274)3.016030+3(3.016030−1.007276)
=0.542
Now, K
2
=
0.375±(0.375
2
+0.542)
1/2
=1.2MeV (- value is invalid because KE could not imaginary number )
So,
K
2
K
1
=
1.2
3
=2.5
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