Question

The half-life of ⁵²Mn is 5.59 days.

(a)

Convert the half-life to units of seconds.

(b)

What is the decay constant (in s⁻¹) for this isotope?

(c)

Suppose a sample of ⁵²Mn has an activity of 0.510 µCi. What is this activity expressed in the SI unit of becquerels (Bq)?

(d)

How many ⁵²Mn nuclei are needed in the sample in part (c) to have the activity of 0.510 µCi?

E. Now suppose that a new sample of ⁵²Mn has an activity of 6.60 mCi at a given time.

How many half-lives will the sample go through in next 40.2 days? (Enter your answer for the number of half-lives to at least one decimal place.)

What is the activity of this sample (in mCi) at the end of 40.2 days?

Answer 1

Use the half life to find the decay constant and hence find the activity as shown below

t/ 11 5.59 days 5.59x 24X 36005 4.83X105 s b) decay constant 2 = luz te/2 - en 2 4.83 X 1055 11 1.44X10-657 1 ci c 3.781020 Bq So, 0. 51 dCi 3.7x1010X0.5IX 10-6 Bq, = 18870 Bq d) activity A = an 18870 Bq = (1.44 x 10-6574) N 1:31X1010 N = e ) no. of half-lives is, m = 40.2 days ts/2 40.2 days 5.59 days 7.2 A = A. e-nt = Now A. - initial activity 6.6mci 2= en 2 ta 40.2 days: ts/2 so/ at In 2 x 40.2 en 2 5.59 X40.2 t1/2 = 4.98 So, A こ Aoerat = = 6.6mcile -6.98) 00452 mci