Question

A freshly isolatedsample of 90Y was found to have an activity of 8.9 x 105 disintegrations per minute at 1:00pm on December 3, 2003. At 2:15pm on Dec, 17, 2003, its activity was redetermined and found to be 2.6 x 104 disintegrations per minute. Calculate the half-life of 90Y.

Answer 1

3960 minutes

Explanation:

From;

0.693/t1/2 = 2.303/t log No/N

t1/2 = half life of 90Y

No = initial activity of 90Y = 8.9 x 10^5 disintegrations per minute

N= activity of 90Y after time t = 2.6 x 10^4 disintegrations per minute

t = time taken

Since the time elapsed is 20,235 minutes, we substitute into the formula above.

0.693/t1/2 = 2.303/20235 log 8.9 x 10^5/2.6 x 10^4

0.693/t1/2 = 1.75 * 10^-4

t1/2 = 0.693/ 1.75 * 10^-4

t1/2 = 3960 minutes