JEE PYQ: Nuclear Physics Question 27
Question 27 - 2019 (10 Jan Shift 2)
Consider the nuclear fission $\text{Ne}^{20} \to 2\text{He}^4 + \text{C}^{12}$
Given that the binding energy/nucleon of $\text{Ne}^{20}$, $\text{He}^4$ and $\text{C}^{12}$ are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement:
(1) energy of 12.4 MeV will be supplied
(2) 8.3 MeV energy will be released
(3) energy of 3.6 MeV will be released
(4) energy of 11.9 MeV has to be supplied
Show Answer
Answer: (4)
Solution
$\text{Ne}^{20} \to 2\text{He}^4 + \text{C}^{12}$ $8.03 \times 20 \quad 2 \times 7.07 \times 4 + 7.86 \times 12$ $\therefore E_B = (BE){\text{react}} - (BE){\text{product}} = 9.72$ MeV therefore, energy has to be supplied
BETA
