Question

Cobalt-60 is radioactive and has a half life of 5.26 years. Calculate the activity of a 4.5 mg sample of cobalt-60. Give your answer in becquerels and in curies. Round your answer to 2 significant digits. Bq ? Ci

Answer 1

time= 5.26 years

= 5.26*3.154*10^7 seconds

= 1.659*10^8 seconds

Given:

Half life = 1.659*10^8 s

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

k = (ln 2) / k

= 0.693/(half life)

= 0.693/(1.659*10^8)

= 4.177*10^-9 s-1

Molar mass of Co = 58.93 g/mol

mass(Co)= 4.5 mg

= 0.0045 g

use:

number of mol of Co,

n = mass of Co/molar mass of Co

=(4.5*10^-3 g)/(58.93 g/mol)

= 7.636*10^-5 mol

use:

number of nuclei = number of moles * Avogadro’s number

= 7.636*10^-5 * 6.022*10^23 nuclei

= 4.599*10^19 nuclei

Activity,

A = k*N

k is decay constant

N is number of Nuclei

now use:

A = k*N

= 4.177*10^-9s-1 * 4.599*10^19 nuclei

= 1.921*10^11decay/s

= 1.921*10^11Bq

= 1.921*10^11/3.7*10^10 Curie

= 5.192 Curie

Answer:

1.9*10^11 Bq

5.2 Ci