2004 AIPMT NucleiMass defect and binding energy Easy Question 4 of 5 in Mass defect and binding energy

$M_P$ denotes the mass of a proton and $M_n$ that of a neutron. A given nucleus, of binding energy B, contains Z protons and N neutrons. The mass M(N, Z) of the nucleus is given by (c is velocity of light)

A $M(N, Z) = NM_n + ZM_P + Bc^2$
B $M(N, Z) = NM_n + ZM_P - \dfrac{B}{c^2}$
C $M(N, Z) = NM_n + ZM_P + \dfrac{B}{c^2}$
D $M(N, Z) = NM_n + ZM_P - Bc^2$

The answer is B.

Solution

The mass of a nucleus is slightly less than the sum of the masses of its constituent nucleons. This mass difference (mass defect) is equivalent to the binding energy.
$\Delta m = \dfrac{B}{c^2} = (ZM_P + NM_n) - M(N, Z)$
Rearranging:
$M(N, Z) = NM_n + ZM_P - \dfrac{B}{c^2}$

Concept Nuclear mass from binding energy

Mass defect and binding energy in past papers

5 questions from this chapter have appeared across 5 exam years.