Physics Circle home Student Login

Home›Chapters›Nuclear Physics PYQs›Mass Defect & Binding Energy›Question 28

NEET 2003 (AIPMT) · Mass Defect & Binding Energy

Mass Defect & Binding Energy question from NEET 2003 (AIPMT)

Q28NEET 2003 · AIPMTMedium

Mass Defect & Binding Energy

The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u = atomic mass unit). The binding energy of 24{}_{2}^{4} He is

(Given helium nucleus mass ≈\approx 4.0015 u.)

Show answer and solution

Answer: (c)

  1. Δm=2(1.0073)+2(1.0087)−4.0015=0.0305\Delta m = 2(1.0073) + 2(1.0087) - 4.0015 = 0.0305 u
  2. BE=0.0305×931≈28.4BE = 0.0305 \times 931 \approx 28.4 MeV
  3. So the answer is (c) 28.4 MeV.
Report a mistake

More questions on Mass Defect & Binding Energy