JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 25)
Consider the nuclear fission
Ne20 $$ \to $$ 2He4 + C12
Given that the binding energy/ nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement -
Ne20 $$ \to $$ 2He4 + C12
Given that the binding energy/ nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement -
8.3 MeV energy will be released
energy of 11.9 MeV has to be supplied
energy of 12.4 MeV will be supplied
energy of 3.6 MeV will be released
Explanation
Ne20$$~ \to $$ 2He4 + C12
Q – value, EB = (BE)react $$-$$ (BE)product
= (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12)
= 9.72 MeV
Q – value, EB = (BE)react $$-$$ (BE)product
= (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12)
= 9.72 MeV
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