Question

A freshly isolated sample of 90Y was found to have an activity of 1.0 × 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 2.7 × 103 disintegrations per minute. Calculate the half-life of 90Y.

Enter your answer in scientific notation.

× 10   minutes

Answer 1

we have:

[A]o = 100000 M

[A] = 2700 M

t = 14 days + 1 hour + 15 minutes

= 14*24*60 + 60 + 15

= 20235 min

use integrated rate law for 1st order reaction

ln[A] = ln[A]o - k*t

ln(2.7*10^3) = ln(1*10^5) - k*2.024*10^4

7.901 = 11.51 - k*2.024*10^4

k*2.024*10^4 = 3.612

k = 1.785*10^-4 min-1

Given:

k = 1.785*10^-4 min-1

use relation between rate constant and half life of 1st order reaction

t1/2 = (ln 2) / k

= 0.693/(k)

= 0.693/(1.785*10^-4)

= 3.882*10^3 min

Answer: 3.88*10^3 min