Q 12-13-108JEE MainJEE Main 2021 (25 Jul, Shift 2)Easy
From the given data, the amount of energy required to break the nucleus of aluminium $^{27}_{13}\text{Al}$ is ______ $x\times10^{-3}$ J
Mass of neutron $= 1.00866$ u
Mass of proton $= 1.00726$ u
Mass of Aluminium nucleus $= 27.18846$ u
(Assume $1$ u corresponds to $x$ J of energy)
(Round off to the nearest integer)
Numerical value type. Enter your answer.
Answer: 27
Mass defect:
$$\Delta m = 13\times1.00726 + 14\times1.00866 - 27.18846 = 13.09438 + 14.12124 - 27.18846 = 0.02716\ \text{u}$$
Energy $= 0.02716x$ J $= 27.16x\times10^{-3}$ J $\approx 27x\times10^{-3}$ J.
Solution by Sreeraj P, M.Sc Physics