From the given data, the amount of energy required to break the nucleus of aluminium _13^27 Al is \underline\hspace2cm x \times 10 - 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)

Atoms and Nuclei PYQ — JEE Main 2021 · 25 Jul · Shift 2 · Q68 with Solution

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Correct answer: 27

The energy required to break a nucleus completely into its constituent nucleons is its binding energy.

For ${}^{27}_{13}\text{Al}$ :

So,

\Delta m = Zm_p + Nm_n - M_{\text{nucleus}}

Given:

m_p = 1.00726\,u, \quad m_n = 1.00866\,u, \quad M_{\text{Al nucleus}} = 27.18846\,u

Mass of 13 protons:

13 \times 1.00726 = 13.09438\,u

Mass of 14 neutrons:

14 \times 1.00866 = 14.12124\,u

Total mass of nucleons:

13.09438 + 14.12124 = 27.21562\,u

\Delta m = 27.21562 - 27.18846 = 0.02716\,u

Since $1\,u$ corresponds to $x$ J of energy,

E = 0.02716\,x \text{ J}

Now write this in the form:

(\text{integer}) \times x \times 10^{-3}\text{ J}

We have,

0.02716\,x = 27.16 \times x \times 10^{-3}\text{ J}

Rounding to the nearest integer:

27.16 \approx 27

The amount of energy required is:

27 \times x \times 10^{-3}\text{ J}

So the required integer is 27.