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

Answer

A, C

Explanation

`t_(1/2) = 10 hrs, A_0 = 1 Ci`
Activity after 9 hrs.
`A = A_(0)e^(-lambdat)`
`= 1 xx e^(-6.693/10) xx 9`
`= e^(-6.693/10) xx 9`
`= 0.5359 = 0.536 Ci`
No. of atoms left after 9^(th) hour,
`A_(g) = lambdaN_(g)`
`N_(g) = A_90)/lambda`
`= (0.536 xx 10 xx 3.7 xx 10^(6) xx 3600)/(0.693)` ltbgt ` = 28.6176 xx 10^(10) xx 3600`
`= 103.023 xx 10^(13)`
Activity after 10 hrs.
A = A_(0)e^(-lambdat)`
` = 1 xx e^(-0.693 /10) xx 100.5 Ci`
No. of atoms left after `10^(th)` hour
`A_(10) = lambdaN_(10)`
`N_(10) = A_(10)/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) xx 10^(13)`
` = 6.92 xx 10^(13)` .