Question

Be sure to answer all parts. A freshly isolated sample of 90Y was found to have an activity of 7.0 × 105 disintegrations per minute at 1:00 p.m. on December 3, 2006. At 2:15 p.m. on December 17, 2006, its activity was measured again and found to be 1.9 × 104 disintegrations per minute. Calculate the half-life of 90Y. Enter your answer in scientific notation.

Answer 1

Time from 1:00 p.m. on december 3, 2006 to 2:15 p.m. om December 2006 is 14 days, 1 hours and 15 minutes.

radioactive decay is given in disintegrations per minute. So, let's calculate the time in minutes.

14 days x (24 hours/1 day) x (60 minutes/1 hour) = 20160 minutes

1 hour x (60 minutes/1 hour) = 60 minutes

Total minutes = 20160 + 60 + 15 = 20235 minutes

Let's represent initial activity as A0 and final activity as A.

A0 = 7.0 x 10^5 disintegrations per minute

A = 1.9 x 10^4 disintegrations per minute

Radioactive decay follows first order kinetics.

lnA = -kt + lnA0

ln(1.9 x 10^4) = - k(20235 min) + ln(7.0 x 10^5)

9.852 = - 20235 min(k) + 13.459

9.852 - 13.459 = - 20235 min(k)

-3.607 = - 20235 min(k)

k = 3.607/20235 min = 1.78 x 10^-4 min^-1

for first order kinetics, half life = 0.693/k

half life = 0.693/(1.78 x 10^-4 min^-1)

half life = 3.89 x 10^3 min