JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Modern Physics

From the given data, the amount of energy required to break the nucleus of aluminium

_{13}^{27} Al

x \times 1 0^{- 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 xJ of energy)
(Round off to the nearest integer)

Answer: 27

Solution

Aluminium-27 contains

13

protons and

14

neutrons, whose separate masses add to

13 \times 1.00726 + 14 \times 1.00866 = 27.21562 u

The mass defect is

Δ m = 27.21562 - 27.18846 = 0.02716

u. Breaking the nucleus into free nucleons costs the energy equivalent of this mass defect, and one unified mass unit is taken to correspond to

x

joule, so the energy required is

0.02716 x J = 27.16 \times 1 0^{- 3} x J

Rounded to the nearest integer, the energy required is

27 \times 1 0^{- 3} x

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Integer answer