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

Answer: (B)

$20 \, \mu \text{g}$

Answer 2

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 $\mu g$