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JEE Main 201910 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 10 January
Shift: Evening Shift
Subject: Physics
Chapter: Nuclear Physics

JEE Main 2019 · 10 January · Evening Shift · Physics · Nuclear Physics

Consider the nuclear fission, Ne ^ 20 → 2 He ^ 4 + C ^ 12 . Given...

Consider the nuclear fission,

{Ne}^{20} \to 2{He}^{4} + {C}^{12}

. Given that the binding energy/nucleon of

{Ne}^{20}, {He}^{4}

and

{C}^{12}

are

8.03MeV, 7.86MeV

, respectively. Identify the correct statement:

Options

AEnergy of $12 . 4MeV$ will be supplied.
BEnergy of $9 . 72MeV$ has to be supplied.
CEnergy of $3 . 6MeV$ will be released.
D $8 . 3MeV$ energy will be released.

Correct answer

B. Energy of $9 . 72MeV$ has to be supplied.

Step-by-step solution

Q = {E}_{f} - {E}_{i}

= \left(12 \times 7.86 + 4 \times 7.07 + 4 \times 7.07\right) - ( 20 \times 8.03 )

Q = - 9.72MeV

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