Question:
Condiser the nuclear fission ${Ne^{20} -> 2 He^{4} + C^{12}} $
Given that the binding energy/nucleon of ${Ne^{20}, He^{4}}$ and ${C^{12}}$ are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement :
Condiser the nuclear fission ${Ne^{20} -> 2 He^{4} + C^{12}} $
Given that the binding energy/nucleon of ${Ne^{20}, He^{4}}$ and ${C^{12}}$ are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement :
Updated On: Apr 3, 2026
- 8.3 MeV energy will be released
- energy of 12.4 MeV will be supplied
- energy of 11.9 MeV has to be supplied
- energy of 3.6 MeV will be released
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The Correct Option is C
Solution and Explanation
${Ne^{20} -> 2He^{4} + C^{12}}$
$8.30 \times 20 \; \; \; \; 2\times 7.07 \times 4 + 7.86 \times 12$
$\therefore \; E_B = (BE)_{react} - (BE)_{product} = 9.72 \, MeV$
$8.30 \times 20 \; \; \; \; 2\times 7.07 \times 4 + 7.86 \times 12$
$\therefore \; E_B = (BE)_{react} - (BE)_{product} = 9.72 \, MeV$
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Concepts Used:
Nuclei
In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.
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Following are the terms related to nucleus:
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