#### Question

The decay constant of `""_80^197`Hg (electron capture to `""_79^197`Au) is 1.8 × 10^{−4} S^{−1}. (a) What is the half-life? (b) What is the average-life? (c) How much time will it take to convert 25% of this isotope of mercury into gold?

#### Solution

Given :-

Decay Constant of `""_80^197"Hg" , lambda = 1.8 xx 10^-4 "s"^-1`

(a)

Half-life, `T_"1/2" = 0.693/lambda`

`⇒ T_"1/2" = 0.693/(1.8 xx 10^-4)`

= 3850 s=64 minutes

(b)

Average life, `T_(av) = T_"1/2"/0.693`

`= 64/0.693`

= 92 minutes

(c)

Number of active nuclei of mercury at t = 0 = N_{0} = 100

Active nuclei of mercury left after conversion of 25% isotope of mercury into gold = N = 75

Now , `N/N_0 = e^(-lambda t)`

Here,

N = Number of inactive nuclei

`N_0` = Number of nuclei at t = 0

`lambda =` Disintegration constant

On substituting the values, we get

`75/100 = e^(-lambdat)`

`⇒ 0.75 = e^(-lambda x)`

`⇒ "In" 0.75 = - lambda t`

`⇒ t = ("In" 0.75)/-0.00018`

= 1600 s