Discussion
When a helium nucleus is formed, there is always some degree of loss of mass. If the loss of mass equals 3.1 x 10–5 kg during the formation of one mol of it, what is the binding energy?
*This question is included in 13. Radioactivity, question #10
| (A) | 3.1 x 10–5 J |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 07/26/2013 18:42
Shouldn't it be c
Posted: 12/14/2015 16:44
Reply: agreed, according to the formula it should be c
Note that the 10^8 is squared, so it becomes 10^16. Multiplying this by 10^-5 yields
10^-5 x 10^16 = 10^11
Now, writing 28 in scientific notation yields 2.8 x 10^1. Multiplying this by 10^11 yields
(2.8 x 10^1)(10^11) = 2.8 x 10^12
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