If we start with 1.000 g of cobalt-60, 0.675 g will remain after 3.00 yr. this means that the of is _____

1 answer:

8 0

Cobalt-60 is undergoing a radioactivity decay.

The formula of the decay is n=N(1/2)∧(T/t).
Where N ⇒ original mass of cobalt
 n ⇒ remaining mass of cobalt after 3 years
T ⇒ decaying period
t ⇒ half-life of cobalt.

So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
3/t= 0.567

t = 3÷0.567
= 5.290626524

the half-life of Cobalt-60 is 5.29 years.

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