JEE Physics 2023 Question with Solution

Given: Nucleus A has $Z = 17$ and equal number of protons and neutrons, with binding energy per nucleon $1.2 \, \text{MeV}$. Nucleus B has $26$ nucleons and binding energy per nucleon $1.8 \, \text{MeV}$.

Find: The difference in total binding energies of B and A.

For nucleus A, number of neutrons equals number of protons, so neutrons $= 17$. Therefore mass number is

$$A = 17 + 17 = 34$$

Total binding energy of nucleus A is

$$BE_A = 1.2 \times 34 = 40.8 \, \text{MeV}$$

For nucleus B, mass number is $26$. Hence total binding energy is

$$BE_B = 1.8 \times 26 = 46.8 \, \text{MeV}$$

Now the difference in binding energies is

$$\Delta BE = BE_B - BE_A = 46.8 - 40.8 = 6 \, \text{MeV}$$

Therefore, the difference in binding energy is $6 \, \text{MeV}$.

Given: Binding energy per nucleon of A is $1.2 \, \text{MeV}$ and of B is $1.8 \, \text{MeV}$.

Find: Difference in total binding energies.

First compute the mass number of A using equal protons and neutrons:

$$A_A = 17 + 17 = 34$$

For B, the mass number is already $26$.

Now multiply binding energy per nucleon by mass number for each nucleus:

$$BE_A = 1.2 \times 34 = 40.8 \, \text{MeV}$$ $$BE_B = 1.8 \times 26 = 46.8 \, \text{MeV}$$

Subtract directly:

$$BE_B - BE_A = 46.8 - 40.8 = 6 \, \text{MeV}$$

Therefore, the required numerical value is $6$.