Q4. A coin of mass 3.0 g is made entirely of copper-63 atoms (62.92960 u). How much nuclear energy would separate all its neutrons and protons? (NCERT Exercise 13.3)
Per nucleus, ΔM = 29 × 1.007825 + 34 × 1.008665 − 62.92960 = 0.591935 u, so Eb ≈ 551.4 MeV. The coin has (3.0/62.92960) × 6.023 × 1023 ≈ 2.87 × 1022 atoms, so the total is about 1.583 × 1025 MeV ≈ 2.53 × 1012 J (NCERT Exercise 13.3, page 321). NCERT’s answer on page 351, 1.584 × 1025 MeV, differs only in the last digit.
Q5. Find the binding energy of the deuteron, 21H (2.014102 u).
ΔM = 1.007825 + 1.008665 − 2.014102 = 0.002388 u, so Eb ≈ 2.22 MeV, only about 1.1 MeV per nucleon: the deuteron is loosely bound (data from NCERT page 321).
Q6. How much mass is converted into energy when 1 MJ is released?
m = E/c2 = 106/(3 × 108)2 ≈ 1.1 × 10−11 kg (NCERT Eq. (13.6), page 310).
Q7. Find the binding energy per nucleon of neon-20 (19.992439 u).
ΔM = 10 × 1.007825 + 10 × 1.008665 − 19.992439 = 0.172461 u, so Eb ≈ 160.6 MeV and Ebn ≈ 8.03 MeV (data from NCERT page 321).
Q8. Why does a nucleon deep inside a large nucleus have about the same binding energy whatever the size of the nucleus?
The nuclear force has a short range, so a nucleon feels only its nearest neighbours; adding more nucleons far away does not change its binding. This is the saturation property (NCERT pages 312–313).
Q9. Find the energy needed to remove a neutron from tritium, 31H (3.016049 u), leaving a deuteron (2.014102 u).
ΔM = 2.014102 + 1.008665 − 3.016049 = 0.006718 u, so Sn ≈ 6.26 MeV (data from NCERT page 321).
Q10. For the rare radiative branch, two deuterons form helium-4 plus a gamma photon (and nuclear recoil). How much energy is released?
ΔM = 2 × 2.014102 − 4.002603 = 0.025601 u, so about 23.8 MeV is shared by photon energy and recoil; the usual deuteron–deuteron branches instead produce tritium plus a proton or helium-3 plus a neutron (data from NCERT page 321).