# If there is a loss of 5.63 x 10^-7 kg of mass in a nuclear reaction, how many Joules of energy would be released?

## Recall that $c = 3 x {10}^{8} \frac{m}{s}$

Sep 14, 2016

Well, $E = m {c}^{2}$, or so the physicists assure us.

#### Explanation:

Thus, $E = 5.63 \times {10}^{-} 7 k g \times {\left(3 \times {10}^{8} \cdot m \cdot {s}^{-} 1\right)}^{2}$

$=$ $5.63 \times {10}^{-} 7 k g \times 9 \times {10}^{16} \cdot {m}^{2} \cdot {s}^{-} 2$

And it is a fact (the which I would check!) that $1 \cdot J$ $=$ $1 \cdot k g \cdot {m}^{2} \cdot {s}^{-} 2$, which are the units we got in the answer.

So we have approx. $50 \times {10}^{9} \cdot J$ $=$ $50 \times {10}^{3} \cdot M J$