QJEEM24Ph14ac: JEE Main 2024 Physics - Class 12 JEEM24Ph14ac Solution: Binding energy of a certain nucleus is . How much is the…

To determine the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus, we need to use the concept of mass-energy equivalence provided by Einstein's famous equation:

$E= \Delta m \cdot c^2$

where:

• E is the binding energy.
• $\Delta m$ is the mass defect (difference in mass).
• $c$ is the speed of light in vacuum, which is approximately $3 \times 10^8 \, \text{m/s}$ .

Given:

• Binding energy, $E = 18 \times 10^8 \, \text{J}$ .

We need to find $\Delta m$ , so rearranging the equation:

$\Delta m = \frac{E}{c^2}$

Substitute the given values:

$\Delta m = \frac{18 \times 10^8}{(3 \times 10^8)^2}$

Calculate the value:

$\Delta m = \frac{18 \times 10^8}{9 \times 10^{16}}$

$\Delta m = \frac{18}{9} \times 10^{-8}$

$\Delta m = 2 \times 10^{-8} \, \text{kg}$

To convert the mass from kilograms to micrograms ( $\mu g$ ), we use the conversion factor $1 \, \text{kg} = 10^9 \, \mu g$ :

$\Delta m = 2 \times 10^{-8} \, \text{kg} \times 10^9 \, \frac{\mu g}{\text{kg}}$

$\Delta m = 2 \times 10^{1} \, \mu g$

$\Delta m = 20 \, \mu g$

Therefore, the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus is 20 $\mu g$ .

Option A (20 $\mu g$ ) is the correct answer.