KCET · Physics · Nuclear Physics
Consider the nuclear fission reaction \({ }_0^1 n+{ }_{92}^{235} \mathrm{U} \longrightarrow{ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+3{ }_0^1 n\). Assuming all the kinetic energy is carried away by the fast neutrons only and total binding energies of \({ }_{92}^{235} \mathrm{U},{ }_{56}^{144} \mathrm{Ba}\) and \({ }_{36}^{89} \mathrm{Kr}\) to be \(1800 \mathrm{MeV}, 1200\) MeV and 780 MeV respectively, the average kinetic energy carried by each fast neutron is (in MeV )
- A \(200\)
- B \(180\)
- C \(67\)
- D \(60\)
Answer & Solution
Correct Answer
(D) \(60\)
Step-by-step Solution
Detailed explanation
Binding energy \({ }_{92}^{235} \mathrm{U}, B E_1=1800 \mathrm{MeV}\)
Binding energy of \({ }_{56}^{144} \mathrm{Ba}, B E_2=1200 \mathrm{MeV}\)
Binding energy of \({ }_{36}^{89} \mathrm{Kr}, B E_3=780 \mathrm{MeV}\)
\(\therefore\) Binding energy of reactants \(=B E_1=1800 \mathrm{MeV}\)
Binding energy of products \(=B E_2=B E_2+B E_3\)
\(=1200+780\)
\(=1980 \mathrm{MeV}\)
\(\therefore\) Average kinetic energy carried by each fast neutron
Binding energy of products
\(=\frac{- \text { Binding energy of reactants }}{3}\)
\(=\frac{1980-1800}{3}=60 \mathrm{MeV}\)
Binding energy of \({ }_{56}^{144} \mathrm{Ba}, B E_2=1200 \mathrm{MeV}\)
Binding energy of \({ }_{36}^{89} \mathrm{Kr}, B E_3=780 \mathrm{MeV}\)
\(\therefore\) Binding energy of reactants \(=B E_1=1800 \mathrm{MeV}\)
Binding energy of products \(=B E_2=B E_2+B E_3\)
\(=1200+780\)
\(=1980 \mathrm{MeV}\)
\(\therefore\) Average kinetic energy carried by each fast neutron
Binding energy of products
\(=\frac{- \text { Binding energy of reactants }}{3}\)
\(=\frac{1980-1800}{3}=60 \mathrm{MeV}\)
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