Question

The binding energy of a certain nucleus is \(18 \times 10^8 \, \text{J}\). How much is the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus?

• A. \(0.2 \, \mu \text{g}\)
• B. \(20 \, \mu \text{g}\)
• C. \(2 \, \mu \text{g}\)
• D. \(10 \, \mu \text{g}\)

Answer 1

The Correct Option is B

Using the mass-energy equivalence \( E = \Delta m c^2 \), the mass defect \( \Delta m \) is calculated as:
\[ \Delta m = \frac{E}{c^2}. \]
Substituting the values:
\[ E = 18 \times 10^8 \, \text{J}, \, c = 3 \times 10^8 \, \text{m/s}, \]
\[ \Delta m = \frac{18 \times 10^8}{(3 \times 10^8)^2} = \frac{18 \times 10^8}{9 \times 10^{16}} = 2 \times 10^{-8} \, \text{kg}. \]
Converting \( \Delta m \) to micrograms (\( \mu g \)):
\[ \Delta m = 2 \times 10^{-8} \, \text{kg} = 20 \, \mu g. \]
Final Answer: 20 \( \mu g \)