JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Modern Physics

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Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments 'B' and 'C' of mass numbers 105 and 115. The binding energy of nucleons in 'B' and 'C' is 6.4 MeV per nucleon. The energy Q released per fission will be :

(A)0.8 MeV
(B)275 MeV
(C)220 MeV
(D)176 MeV✓ Correct

Answer: (D)

Solution

The energy released in fission is the difference between the total binding energy of products and reactants:

Q = (B.E.)_{products} - (B.E.)_{reactant}

Q = (105 + 115) \times 6.4 - 220 \times 5.6

Q = 220 \times 6.4 - 220 \times 5.6 = 220 (6.4 - 5.6) = 220 \times 0.8 = 176

MeV

The answer is (D) 176 MeV.

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option

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