On your birthday, you measure the activity of the sample `""^(210)Bi` which has a half - life of 5.01 days. The initial activity that you measure is `1muCi`. (a) What is the approximate activity of the sample on your next birthday? Calculate (b) the decay constant (c) the mean life (d) initial number of atoms.

Explanation

Half life of `""^(210)Bi" "T_(1//2)=5.01` days
Decay constant `lamda=?`
Mean life `tau=?`
(a) Given `T_(1//2)` = 5.01days
converting to seconds
`T_(1//2)=5.01xx24hrsxx3600sec`.
`T_(1//2)` = 432,864 S
We know that `T_(1//2)=0.693/lamda`
`rArr=0.693/T_(1//2)=(0.693)/(432,864)=1.601xx10^(-6)s^(-1)`
Decay constant `lamda=1.601xx10^(-6)s^(-1)`
Activity of the sample
`R=R_(0)e^(-lamdat)`
After one year t = `365xx24xx3600=3.15xx10^(7)`
Given `R_(0)=1muCi`
`rArrR=1xx10^(-6)e^(-(1.6xx10-6)(3.15xx107))`
`rArrR=1.29xx10^(-22)muCi`
(b) Decay constant `lamda=(0.1931)/T_(1//2)`
`T_(1//2)=5.01xx60xx60xx12=432864`sec
`lamda=0.6931/432864=1.6xx10^(-6)s`.
(c) Mean life `tau=1/lamda`
`lamda=0.6931/5.01=0.1383`
`tau=1/0.1383=7.24` days
(d) Number of atoms initially `N_(0)=R_(0)/lamda`
`N_(0)" given "=1muCi` Convert to Becq by multiplying by `3.7xx10^(10)`
`N_(0)=(1xx10^(-6)xx3.7xx10^(10))/(1.601xx10^(-6))`
`N_(0)=2.31xx10^(10)`