question_answer

Consider the nuclear fission \[N{{e}^{20}}\to 2H{{e}^{4}}+{{C}^{\text{12}}}\] Given that the binding energy/nucleon of\[N{{e}^{20}}\], \[H{{e}^{4}}\] and \[{{C}^{12}}\] are respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement- [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

A) 8.3 MeV energy will be released

B) energy of 11.9 MeV has to be supplied

C) energy of 12.4 MeV will be supplied

D) energy of 3.6 MeV will be released

Correct Answer: B

Solution :

\[{{E}_{supplied}}={{E}_{1}}=20\times B{{E}_{N}}\,of\,\,N{{e}^{20}}\] \[{{E}_{released}}=12\,\,B{{E}_{N}}\,of\text{ }{{C}^{12}}+2\times 4\times \,B{{E}_{N}}\,of\text{ }H{{e}^{4}}\] Supplied energy \[=20\,B{{E}_{N}}\,of\text{ }N{{e}^{20}}-12\times B{{E}_{N}}\] \[of\text{ }{{C}^{12}}-2\,\,\times \,\,4\times B{{E}_{N}}\,of\text{ }H{{e}^{4}}=11.9\text{ }MeV\]