Question

The half-life of arsenic-81 is 33 seconds. Suppose you have 100 g - how much will remain undecayed after 12.5 days?

Answer 1

The Rutherford-Soddy law of radioactive decay is :
`N(t)=N_0 e^(-\lambdat); \qquad \lambda = \ln(2)/t_{1/2}`.

Given : `t_{1/2} = 33 sec; => \lambda = 2.1004\times10^{-2} sec^{-1}`

`t=12.5 days=1.08\times10^6 sec;`
`(N(t))/N_o = e^(-\lambdat) = 0.3679\times10^{-5}`

`N_o-` Number of As-81 nuclei at the beginning (`t=0`),
`N(t) -` Number of As-81 nuclei after time `t`,
`M_o-` Mass of As-81 at the beginning (`t=0`)
`M(t) -` Mass of As-81 nuclei after time `t`,

`(M(t))/M_o = (N(t))/N_o = 0.3679\times10^{-5};`
`M(t) = 0.3679\times10^{-5}\timesM_o=0.3679` milli-grams