Question

after one hour, 1/16 of the initial amount of certain radioactive isotope remains undecayed. the half life of the isotope is:

Answer 1

(1/2)^x=1/16
ln 1/2^x=x ln 1/2=ln 1/16
x=4
4 half-lives in one hour=1/4 hour for the half-life of the isotope
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Answer 2

`t_(1/2) = 0.25 hours`

Step-by-step explanation:

As we know that

`N = N_o e^(-\lambda t)`

here we know that

`\lambda = (ln2)/(t_(1/2))`

so we will have

`N = N_o e^{(t ln2)/(t_(1/2))}`

so it will be

`N = N_o((1)/(2))^{(t)/(t_(1/2))}`

now we know that

`t = 1 hour`

`N = (N_o)/(16)`

`(1)/(16) = ((1)/(2))^{(t)/(t_(1/2))}`

`4 = (1)/(t_(1/2))`

`t_(1/2) = 0.25 hours`