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Question

Match the nuclear processes given in column I with the appropriate option(s) in column II

Column I
Column II
(A) Nuclear fusion (P) Absorption of thermal neutrons by 23592U
(B) Fission in a nuclear reactor (Q) 6027Co nucleus
(C) β-decay
(R)
Energy production in stars via hydrogen
conversion to helium
(D) γ-ray emission (S) Heavy water
(T) Neutrino emission

A
(A)(R,S);(B)(T,S);
(C)(P,Q,R,T);(D) \rightarrow (P,Q,T)$
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B
(A)(R);(B)(P,S);
(C)(P,Q,R,T);(D) \rightarrow (P,Q,R,T)$
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C
(A)(R,T);(B)(P,S);
(C)(P,R,T);(D) \rightarrow (P,Q,S,T)$
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D
(A)(Q,T);(B)(P,);
(C)(P,R,T);(D) \rightarrow (P,Q,S,T)$
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Solution

The correct option is B (A)(R);(B)(P,S);
(C)(P,Q,R,T);(D) \rightarrow (P,Q,R,T)$
for option (A):
Nuclear fussion
- Absorption of thermal neutrons by 23592U is only in nuclear fission. So (P) this is wrong.
6027Co nucleus having β decay so no nuclear fussion. So option (Q) is also wrong.
Energy production in stars occurs via hydrogen conversion to helium and thus it is related to nuclear fussion. So option (R) is correct.
Heavy water work as a moderator in reactors. So no relation with nuclear fussion.
Neutrino emission is related to β decay, So no relation with nuclear fussion.
Hence, AR

for option (B):
Fission in a nuclear reactor
Absorption of thermal neutrons by 23592U its a part of fission.
Heavy water used in a process of fission.
Hence, BT, S

For option C:
β-decay
-Absorption of thermal neutrons by 23592U, it produces the β-decay.
6027Co nucleus, produces the β-decay.
Energy production in stars via hydrogen conversion to helium it devlopes β-decay.
Neutrino emission are the part of β-decay.
Hence, CP,Q,R,T

for option (D):
γ-ray emission
-Absorption of thermal neutrons by 23592U doesn't produces the γ-ray. But after this process when fission came γ-ray produces.
6027Co nucleus produces the β-decay, in a condition of β-decay always γ-ray emits.
In fussion γ-ray produces, so option (R) matches.
In neutrino emission γ-ray produces, so option (T) matches.
Hence,
(A)(R)(B)(P,S)
(C)(P,Q,R,T)
(D)(P,Q,R,T)

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