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The radioactive substance Iridium-192 has a half-life of 73.83 days. How many days would it take for a sample of Iridium-192 to decay to 10% of its original amount? Round your answer to the nearest integer.

 Apr 10, 2019

Best Answer 

 #1
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10% =100% * (0.5)^(d/73.83), solve for d
0.1 =0.5^(d/73.83)     Take the log of both sides
log(0.1) =d/73.83 * log(0.5)
d/73.83 =log(0.1) / log(0.5)
d/73.83 =3.3219281.....
d =73.83 * 3.3219281...
d =~245 - days for given amount to decay to 10%.

 Apr 10, 2019
 #1
avatar
+1
Best Answer

10% =100% * (0.5)^(d/73.83), solve for d
0.1 =0.5^(d/73.83)     Take the log of both sides
log(0.1) =d/73.83 * log(0.5)
d/73.83 =log(0.1) / log(0.5)
d/73.83 =3.3219281.....
d =73.83 * 3.3219281...
d =~245 - days for given amount to decay to 10%.

Guest Apr 10, 2019

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