What is the half-life of an isotope that decays to 6.25% of its original activity in 18.9 hours?
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Answer:
Explanation:
Let assume that gas behaves ideally and experiments an isobaric and isothermal processes. The following relationship is applied to determined the final volume:
a) To find the mass after t years:
we will use this formula:
A = Ao / 2^n
when A =the amount remaining
and Ao = the initial amount
and n = t / t(1/2)
by substitution:
∴ A = 200 mg/ 2^(t/30y)
b) Mass after 90 y :
by using the previous formula and substitute t by 90 y
A = 200mg/ 2^(90y/30y)
∴ A = 25 mg
C) Time for 1 mg remaining:
when A= Ao/ 2^(t/t(1/2)
so, by substitution:
1 mg = 200 mg / 2^(t/30y)
∴2^(t/30y) = 200 mg by solving for t
∴ t = 229 y
we will use this formula:
A = Ao / 2^n
when A =the amount remaining
and Ao = the initial amount
and n = t / t(1/2)
by substitution:
∴ A = 200 mg/ 2^(t/30y)
b) Mass after 90 y :
by using the previous formula and substitute t by 90 y
A = 200mg/ 2^(90y/30y)
∴ A = 25 mg
C) Time for 1 mg remaining:
when A= Ao/ 2^(t/t(1/2)
so, by substitution:
1 mg = 200 mg / 2^(t/30y)
∴2^(t/30y) = 200 mg by solving for t
∴ t = 229 y