Question

question 5 1 pts during a nuclear process, an atom has a mass defect of 19.98x10^-17 kg. how much energy (in joules) is released? report to the tenths place. question 6 1 pts a nuclear process releases 1.70x10^14 j of energy. how many grams is the mass defect? round your answer to the hundredths place.

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: Solve for Question 5

Given $m = 19.98\times10^{-17}\ kg$, substitute into $E = mc^{2}$.
$E=(19.98\times 10^{-17})\times(3\times 10^{8})^{2}$
$E=(19.98\times 10^{-17})\times9\times 10^{16}$
$E = 19.98\times9\times10^{-17 + 16}$
$E=179.82\times10^{-1}$
$E = 17.982\ J\approx18.0\ J$

Step3: Solve for Question 6

First, from $E = mc^{2}$, solve for $m$. We get $m=\frac{E}{c^{2}}$. Given $E = 1.70\times10^{14}\ J$ and $c = 3\times10^{8}\ m/s$.
$m=\frac{1.70\times 10^{14}}{(3\times 10^{8})^{2}}$
$m=\frac{1.70\times 10^{14}}{9\times 10^{16}}$
$m=\frac{1.70}{9}\times10^{14 - 16}$
$m=\frac{1.70}{9}\times10^{-2}\ kg$
To convert to grams, multiply by 1000. $m=\frac{1.70\times10^{-2}\times1000}{9}\ g=\frac{17}{9}\ g\approx1.89\ g$

Answer:

Question 5: 18.0
Question 6: 1.89