Question

1. (6) during a nuclear process, the mass defect is calculated to be 2.45x10 - 5 g. how many megajoules of energy is released during this process? 4. (3) if a nuclear reaction produces 1.75 gigajoules of energy, what was the mass defect in milligrams?

Explanation:

Step1: Recall Einstein's mass - energy equivalence formula

$E = mc^{2}$, where $E$ is energy, $m$ is mass, and $c$ is the speed of light ($c = 3\times10^{8}\ m/s$).

Step2: For problem 3

First, convert the mass defect $m = 2.45\times10^{-5}\ g$ to $kg$. Since $1\ g=10^{- 3}\ kg$, then $m = 2.45\times10^{-5}\times10^{-3}\ kg=2.45\times10^{-8}\ kg$. Then use $E = mc^{2}$. Substitute $m = 2.45\times10^{-8}\ kg$ and $c = 3\times10^{8}\ m/s$ into the formula: $E=(2.45\times10^{-8})\times(3\times10^{8})^{2}$. Calculate $(3\times10^{8})^{2}=9\times10^{16}$, then $E = 2.45\times10^{-8}\times9\times10^{16}=22.05\times10^{8}\ J$. Convert to megajoules: $1\ MJ = 10^{6}\ J$, so $E=\frac{22.05\times10^{8}}{10^{6}}=2205\ MJ$.

Step3: For problem 4

We know $E = 1.75\ GJ=1.75\times10^{9}\ J$ and $E = mc^{2}$. Rearrange the formula to solve for $m$: $m=\frac{E}{c^{2}}$. Substitute $E = 1.75\times10^{9}\ J$ and $c = 3\times10^{8}\ m/s$ into the formula: $m=\frac{1.75\times10^{9}}{(3\times10^{8})^{2}}=\frac{1.75\times10^{9}}{9\times10^{16}}=\frac{1.75}{9}\times10^{-7}\ kg$. Convert to milligrams: Since $1\ kg = 10^{6}\ mg$, then $m=\frac{1.75}{9}\times10^{-7}\times10^{6}\ mg=\frac{1.75}{9}\times10^{-1}\ mg\approx0.0194\ mg$.

Answer:

1. 2205 megajoules
2. Approximately 0.0194 milligrams