The half-life of a radioisotope is 10 h. Find the total number of disintegration in the tenth hour measured from a time when the activity was 1 Ci.

#### Solution

Given:

Half-life of radioisotope,`T_(1"/"2)` = 10 hrs

Initial activity, `A_0` = 1 Ci

Disintegration constant, `lambda = 0.693/(10 xx 3600) "s"^-1`

Activity of radioactive sample,

`A = A_0e^(-lambdat)`

Here,* **A*_{0} = Initial activity

`lambda` = Disintegration constant*t* = Time taken

After 9 hours,

Activity , `A = A_0e^(-lambdat) = 1 xx e^(-0.693/(10 xx 3600) xx 9 = 0.536` Ci

`therefore` Number of Atoms left , N = `A/lambda = (0.536 xx 10 xx 3.7 xx 10^10 xx 3600)/0.693 = 103.023 xx 10^13` After 10 hrs

Activity , A " `= A_0e^(-lambdat)`

= `1 xx e^(-0.693/10 xx 10) = 0.5` Ci

Number of atoms left after the 10^{th} hour (`N` ")will be

A" = `lambdaN` "

N" = A"/`lambda`

= `(0.5 xx 3.7 xx 10^10 xx 3.600)/(0.693"/"10)`

= `26.37 xx 10^10 xx 3600 = 96.103 xx 10^13`

Number of disintegrations = (103.023 − 96.103) × 10^{13}

= 6.92 × 10^{13}