CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In an agricultural experiment, a solution containing 1 mole of a radioactive material (t12, = 14.3 days) was injected into the roots of a plant. The plant was allowed 70 hours to settle down and then activity was measured in its fruit, if the activity measured was 1μCi, what per cent of activity is transmitted from the root to the fruit in steady state?


Solution

Given n = 1 mole = 6×1023 atoms, 

t12 = 14.3 days, 

t= 70 hrs. 

dNdt (after time t) = λN

N = N0eλt

= 6×1023×e0.693×70143×24

= 6×1023×0.868

= 5.209×1023

dNdt=5.209×1023×0.69314.3×24

= 0.0105×10233600dishr

= 2.9×105×1023dissec

= 2.9×1017dissec.

Fraction of activity transmitted 

= (1μCi2.9×1017)×100 %

=[(1×3.7×1042.9×1017)×100] %

= 1.275×1011 %

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...


Same exercise questions
QuestionImage
QuestionImage
View More...


People also searched for
QuestionImage
QuestionImage
View More...



footer-image