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Test Index
Atoms and Nuclei Part 3
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Section:
Physics
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© examsnet.com
Question : 80 of 98
Marks:
+1
,
-0
The explosive in a Hydrogen bomb is a mixture of
1
2
H
,
1
3
H
{}_1^2\mathrm{H},{}_1^3\mathrm{H}
1
2
H
,
1
3
H
and
3
6
L
i
{}_3^6\mathrm{Li}
3
6
Li
in some condensed form. The chain reaction is given by
3
6
L
i
+
0
1
n
→
2
4
H
e
+
1
3
H
{}_3^6\mathrm{Li} + {}_0^1\mathrm{n} \rightarrow {}_2^4\mathrm{He} + {}_1^3\mathrm{H}
3
6
Li
+
0
1
n
→
2
4
He
+
1
3
H
1
2
H
+
1
3
H
→
2
4
H
e
+
0
1
n
{}_1^2\mathrm{H} + {}_1^3\mathrm{H} \rightarrow {}_2^4\mathrm{He} + {}_0^1\mathrm{n}
1
2
H
+
1
3
H
→
2
4
He
+
0
1
n
During the explosion the energy released is approximately
[Given :
M
(
L
i
)
=
6.01690
amu
.
M
(
1
2
H
)
=
2.01471
M(\mathrm{Li}) = 6.01690\ \text{amu}. M({}_1^2\mathrm{H}) = 2.01471
M
(
Li
)
=
6.01690
amu
.
M
(
1
2
H
)
=
2.01471
amu.
M
(
2
4
H
e
)
=
4.00388
amu
M({}_2^4\mathrm{He}) = 4.00388\ \text{amu}
M
(
2
4
He
)
=
4.00388
amu
, and
1
amu
=
931.5
1\ \text{amu} = 931.5
1
amu
=
931.5
MeV
]
\text{MeV}]
MeV
]
[29-Jan-2024 Shift 1]
28.12
MeV
28.12\ \text{MeV}
28.12
MeV
12.64
MeV
12.64\ \text{MeV}
12.64
MeV
16.48
MeV
16.48\ \text{MeV}
16.48
MeV
22.22
MeV
22.22\ \text{MeV}
22.22
MeV
Validate
Solution:
👈: Video Solution
3
6
L
i
+
0
1
n
⟶
2
4
H
e
+
1
3
H
\;{}_3^6\mathrm{Li} + {}_0^1\mathrm{n} \longrightarrow {}_2^4\mathrm{He} + {}_1^3\mathrm{H}
3
6
Li
+
0
1
n
⟶
2
4
He
+
1
3
H
1
2
H
+
1
3
H
⟶
2
4
H
e
+
0
1
n
\;{}_1^2\mathrm{H} + {}_1^3\mathrm{H} \longrightarrow {}_2^4\mathrm{He} + {}_0^1\mathrm{n}
1
2
H
+
1
3
H
⟶
2
4
He
+
0
1
n
_____________________
3
6
L
i
+
1
2
H
⟶
2
(
2
4
H
e
)
\;{}_3^6\mathrm{Li} + {}_1^2\mathrm{H} \longrightarrow 2({}_2^4\mathrm{He})
3
6
Li
+
1
2
H
⟶
2
(
2
4
He
)
_____________________
Energy released in process
\;\;\text{Energy released in process}\;
Energy released in process
Q
=
Δ
m
c
2
\;Q = \Delta m c^{2}
Q
=
Δ
m
c
2
Q
=
[
M
(
L
i
)
+
M
(
1
2
H
)
−
2
×
M
(
2
4
H
e
)
]
\;Q = [M(\mathrm{Li}) + M({}_1^2\mathrm{H}) - 2 \times M({}_2^4\mathrm{He})]
Q
=
[
M
(
Li
)
+
M
(
1
2
H
)
−
2
×
M
(
2
4
He
)]
×
931.5
MeV
\times 931.5\ \text{MeV}
×
931.5
MeV
Q
=
[
6.01690
+
2.01471
−
2
×
4.00388
]
\;Q = [6.01690 + 2.01471 - 2 \times 4.00388]
Q
=
[
6.01690
+
2.01471
−
2
×
4.00388
]
×
931.5
MeV
\times 931.5\ \text{MeV}
×
931.5
MeV
Q
=
22.216
MeV
\;Q = 22.216\ \text{MeV}
Q
=
22.216
MeV
Q
=
22.22
MeV
\;Q = 22.22\ \text{MeV}
Q
=
22.22
MeV
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