Question

The unstable isotope ^{234}Th decays by beta emission
with a half-life of 24.5 days.

a) What mass of ^{234}Th will produce
9.9x10^{16} decays per second? (Note: 1u =
1.66x10^{-27} kg). Answer:1.2x10^{-4}kg

b) If the initial decay rate of the sample is
9.9x10^{16} decays per second, what is the decay rate after
59 days? Answer: 1.4x10^{13}kg decays/second

Must show all work to get to the answers I provided. Thank you!

Answer #1

given

234 Th

half life, T' = 24.5 days

a. let there be No atoms of Th

then

activity = lambda*No

where

lambda = ln(2)/T' = ln(2)/24.5 days = ln(2)/24.5*24*60*60 s = 3.2745*10^-7 per s

hence

A = 9.9*10^16 decays per second

so,

9.9*10^16 = 3.2745*10^(-7)*No

No = 3.02335*10^23

hecne moles of atoms = 0.502 moles

molar mass of 234 Th = 234.0436 g

hence

mass of Th required = 234.0436*0.502 = 117.502 g

b. Ao = 9.9*10^16 decays per sec

after 59 days

A = Ao*e^(-lambda*t)

A = 9.9*10^16*e^(-ln(2)*59/24.5)

A = 1.865*10^16 decays per sec

the answer provided is incorrect by intuition as the activity must fall to half after one half life, so after about 49 days, the activity must be 1/4th of given activity, i.e. about 2.5*10^16 decays per second, hence it cannot be 1.4*10^13 decays per second in the next few days

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