Question

pd-100 has a half-life of 3.6 days. if one had 6.02 x 1023 atoms at the start, how many atoms would be present after 20.0 days?

Answer 1

half-life means you divide by 2 
21.6 / 3.6 = 6 hf (hf = half-life) 
then you divide by 2^6 
6.02 x 10^23 / 2^6 = 9.40625 x 10^21 atoms 
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26 
3hf ===> 2^3 
10.0 / 2^3 = 1.25 gram (remaining) 
answer : 10 - 1.25 = 8.75g 
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27 
128 / 2^n = 2 
64 = 2^n 
n = 6 
24 /6 = 4  
4 days is the half-live of the sample 
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28 
2.5/4.46 of 50% initial mass 
(2.5/4.46) x 0.5 x 2 = 0.5605 g 
answer: 2 - 0.5605 = 1.4395 g 
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29 
2^7 = 128 
so 7 half-lifes 
7* 8040 = 56200 days 
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30 
the time of 10 half-lifes 
= 10 x 0.334s = 3.34s