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Question
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, A0 = 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 10th 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) × 1013
= 6.92 × 1013
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