`14 g` of a radioactive substance decays to `7 g` in 20 min. Will the time required be more or less the following processes:
i. `20 g` decreases by `8 g`
ii. `20 g` decreases to `8 g`. Explain

Explanation

Since `14 g` decays to `7 g`d in 20 min, this means half-life period `(t_(1//2)) = 20` min.
In case (i), `20 g` decreases by `8 g`. This means that less than half of the substance decays. Hence, time taken will be less than 20 min.
In ccase (ii), `20 g` decreases to `8 g`. This meansd `12 g` of the substance decays which is mored than half of the original substance. Hence, the time taken will be more than 20 min. The exact calculation of timesdf in the two cases can be done as follows:
Form the first data
` K = (0.693)/(t_(1//2)) = (0.693)/(20) = 0.03465 "min"^(-1)`
In case (i)
`t = (2.303)/(K) "log" (a)/(a - x) = (2.303)/(0.03465) " log" (20)/(20 - g) = 14.74` min
In case (ii)
`t = (2.303)/(K) "log" (a)/(a - x) = (2.303)/(0.03465) "log" (20)/(8) = 26.45` min