Nuclear Chemistry and RadioactivityhardNUMERICAL

The mass of a helium atom (mass number 4) is amu, while the masses of a free neutron and proton are Nuclear Chemistry and Radioactivity Chemistry Question

Question

The mass of a helium atom (mass number 4) is $4.0026$ amu, while the masses of a free neutron and proton are $1.0087$ amu and $1.0078$ amu respectively on the same scale. Assuming the mass of electrons is negligible here for mass defect calculation, the binding energy per nucleon in the helium atom is nearly how many MeV? (Use $1 \text{ amu} \approx 931.5 \text{ MeV}$ and round to the nearest integer).

Answer: 7

💡 Solution & Explanation

Helium (${}_{2}^{4}He$) has 2 protons and 2 neutrons. Expected mass = $2 \times 1.0078 + 2 \times 1.0087 = 2.0156 + 2.0174 = 4.0330$ amu. Mass defect ($\Delta m$) = $4.0330 - 4.0026 = 0.0304$ amu. Total Binding Energy = $0.0304 \times 931.5 \approx 28.317$ MeV. Binding energy per nucleon = $28.317 / 4 \approx 7.07$ MeV. Nearest integer is 7.

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full JEE Mains Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry