Consider the nuclear fission Ne20 → 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 - (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

Answer: (b). BE of ²⁰Ne = 20×8.03 = 160.60 MeV. BE of products = 2×4×7.07 + 12×7.86 = 56.56 + 94.32 = 150.88 MeV. Q = 150.88 - 160.60 = -9.72 MeV, so energy must be supplied. The nearest 'supplied' option is (b) 11.9 MeV.

Consider the nuclear fission Ne20 → 2He4 + C12 Given that the binding… | JEE Main 2019 (10 Jan, Shift 2)

Q23JEE Main 2019 · 10 JAN · SHIFT 2Medium

Nuclear Fission & Fusion

Consider the nuclear fission

Ne20 $\rightarrow$ 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 -

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Answer: (b)

BE of ${}^{20}\text{Ne} = 20\times8.03 = 160.60\text{ MeV}$ .
BE of products $= 2\times4\times7.07 + 12\times7.86 = 56.56 + 94.32 = 150.88\text{ MeV}$ .
$Q = 150.88 - 160.60 = -9.72\text{ MeV}$ , so energy must be supplied.
The nearest 'supplied' option is (b) $11.9\text{ MeV}$ .

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