This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: SOLUTIONS TO MIDTERM II; MAT 127 (1) (20 points total) After 3 days a sample of radon222 decayed to 58% of its original amount. (a) (10 points) What is the halflife of radon222? Solution: Let y ( t ) denote the amount of radon222 at time t . Then y ( t ) = ce kt where c denotes the amount at time t = 0. The “halflife” is the the time t it takes to cut in half the original amount of radon222; so the halflife t is the solution to the equation ( i ) ce kt = c/ 2 . To solve this equation divide both sides by c , then apply the natural log to both sides, and finally divide each side by k to get that ( ii ) t = ln (1 / 2) /k . This gives the half life “in terms of k ”; so to find the half life it remains to compute k . To solve for k we use the information tht the amount of radon222 left after 3 days ( y (3) = ce 3 k ) is 58% of of the initial amount (.58c). We thus have the equation ( iii ) ce 3 k = . 58 c . Dividing this equation by c , applying the natural log to each side, and then dividing by 3 gives ( iv ) k = ln ( . 58) / 3 . Now combining forulae (ii),(iv) gives the following forula for the halflife: ( v ) t = 3 ln (1 / 2) /ln ( . 58) . (b) (10 points) How long would it take the sample to decay to 10% of its original amount? Solution: We are looking for the time t which satisfies ( vi ) ce kt = . 1 c ....
View
Full Document
 Fall '07
 GuanYuShi
 HalfLife, m'th partial sum

Click to edit the document details