What is the `alpha` - activity in disintigration per minute 1 gm sample of `""^(226)Ra. (t_(1//2) = 1620 " year" )`

Explanation

As the rate of disintegration is to be calculated as disintegrations per minute, `t_((1)/(2))` has also to be taken in minutes.
`t_((1)/(2))=1620 xx 365 xx 24 xx 60` minutes
Mole of `""^(226)Ra = (0.001)/(226)`
`therefore` no. of nuclei of `""^(226)Ra`= mole `xx` Av. const
`=(0.001)/(226)xx 6.022 xx 10^(23)`
Now, we have, `-(d(N))/(dt)= lamda (N)= (0.6932)/(t_((1)/(2)))xx N`
`=(0.6932)/(1620 xx 365 xx 24 xx 60) xx (0.001 xx 6.022 xx 10^(23))/(226)`
`=2.16 xx 10^(19)` events/minute