Question

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A radioactive sample is measured to have 4.0x10^6 decay/s. After 6.0 s the decay rate has decreased to 5.0x10^5 decay/s. Determine the lifetime tau and the half-life expressed in seconds. Determine the activity after 20 seconds

Answer 1

Note that

A = Ao e^(-t/T)

--> T = -t / ln (A/Ao)

for exponential decay where A is the amount at time t.
As

A = 5.0E5 decay/s
Ao = 4.0E6 decay/s
t = 6.0 s
T = mean life time

Plugging these values in,

T = 2.88 s   [mean lifetime]

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Note that

Half life = T ln 2

Thus,

Half life = 2.0 seconds   [ANSWER]

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At t = 20 s,

A = Ao e^(-t/T)

= 5.0E5 e^(-20/2.88)

= 488 decay/s   [ANSWER]

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