m = | m_{o} ---------------- [(1 - (^{v}/_{c})^{2}]^{1/2} |
---|
Einstein further showed that the increase in kinetic energy DE is equal to Dm c^{2}.
Example 1
Solution
Discussion
The fraction of mass lost = 2.7e-9 / 18
= 1.5e-10
= 15 ppb (part per billion)
About 3 ng is insignificant (1.5e-10) in a weight of 18 g.
Example 2
Solution
First of all, the mass equivalence of the energy released is
Total mass of product = (8.35272e-24 - 3.133e-26) g
= 8.35269e-24 g
Discussion
The fraction of mass lost = 3.133e-26/8.35272e-24
= 0.38 %
In terms of round figures, about 4 thousandth of the mass is lost.
E-mail: cchieh@uwaterloo.ca