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Physics · 2024 · 2 marks
CBSE 2024 · Region 2 · Set 1 · Q20
The carbon isotope _6^12 C has a nuclear mass of 12.000000 u. Calculate the binding energy of its nucleus. Given m_p = 1.007825 u; m_n = 1.008665 u.
The carbon isotope $\displaystyle {}_{6}^{12}\mathrm{C}$ has a nuclear mass of $\displaystyle 12.000000$ u. Calculate the binding energy of its nucleus. Given $\displaystyle m_p = 1.007825$ u; $\displaystyle m_n = 1.008665$ u.
Marking-scheme solution
Binding Energy $\displaystyle =\left(Z m_{p}+(A-Z) m_{n}-M_{N}\right) \times 931.5 \mathrm{MeV}$
B.E. $\displaystyle =(6 \times 1.007825+6 \times 1.008665-12.000000) \times 931.5 \mathrm{MeV}$
$\displaystyle =(0.09894) \times 931.5 \mathrm{MeV}$
B.E. $\displaystyle =92.16 \mathrm{MeV}$
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CBSE Class 12 Physics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.