The atomic mass of ₆C¹² is 12.000000 u and that of ₆C¹³ is 13.003354 u. The required energy to remove a neutron from ₆C¹³, if mass of neutron is 1.008665 u, will be: (a) 62.5 MeV (b) 6.25 MeV (c) 4.95 MeV (d) 49.5 MeV

Answer: (c). Removing a neutron: ¹³C → ¹²C + n Δ m = m(¹²C) + mn - m(¹³C) = 12.000000 + 1.008665 - 13.003354 = 0.005311 u E = 0.005311 × 931.5 ≈ 4.95 MeV So the answer is (c) 4.95 MeV.

The atomic mass of ₆C¹² is 12.000000 u and that of ₆C¹³ is 13.003354… | JEE Main 2024 (27 Jan, Shift 2)

Q11JEE Main 2024 · 27 JAN · SHIFT 2Medium

Mass Defect & Binding Energy

The atomic mass of ${}_{6}\text{C}^{12}$ is $12.000000\text{ u}$ and that of ${}_{6}\text{C}^{13}$ is $13.003354\text{ u}$ . The required energy to remove a neutron from ${}_{6}\text{C}^{13}$ , if mass of neutron is $1.008665\text{ u}$ , will be:

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Answer: (c)

Removing a neutron: ${}^{13}\text{C} \rightarrow {}^{12}\text{C} + n$
$\Delta m = m({}^{12}\text{C}) + m_n - m({}^{13}\text{C}) = 12.000000 + 1.008665 - 13.003354 = 0.005311$ u
$E = 0.005311 \times 931.5 \approx 4.95$ MeV
So the answer is (c) 4.95 MeV.

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Mass Defect & Binding Energy question from JEE Main 2024 (27 Jan, Shift 2)

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