The Binding energy per nucleon of the nucleus \( {}^{56}_{26}\mathrm{X} \) in units of MeV is:
(Given that \( m_{\mathrm{H}} = 1.007825\,\mathrm{u} \), \( m_{\mathrm{n}} = 1.008665\,\mathrm{u} \) and mass of \( {}^{56}_{26}\mathrm{X} = 55.934939\,\mathrm{u} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8.8\,\mathrm{MeV} \)
The binding energy is the energy equivalent of the mass defect, the difference between the mass of the separate nucleons and the mass of the nucleus. The given hydrogen mass \( m_{\mathrm{H}} \) includes one electron, and so does the atomic mass of X, so the electron masses cancel. For \( Z = 26 \) and \( A = 56 \) there are 26 protons and \( 56 - 26 = 30 \) neutrons.
\( \Delta m = Zm_{\mathrm{H}} + (A - Z)m_{\mathrm{n}} - M \)
\( \Delta m = 26\times1.007825 + 30\times1.008665 - 55.934939 \)
\( \Delta m = 26.203450 + 30.259950 - 55.934939 = 0.528461\,\mathrm{u} \)
\( E_b = 0.528461\times931.5\,\mathrm{MeV} = 492.3\,\mathrm{MeV} \)
\( \displaystyle \frac{E_b}{A} = \frac{492.3\,\mathrm{MeV}}{56} = 8.8\,\mathrm{MeV} \)
Shortcut: Mass number 56 with \( Z = 26 \) is iron, which sits at the top of the binding-energy-per-nucleon curve at about 8.8 MeV per nucleon. Only Option (1) is close to this value; the others are far too large for any nucleus.
Note: NTA picked this question from an NCERT (2026-27) exercise and added the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
