Question

A device used in radiation therapy for cancer contains 1.24 g of cobalt 2760Co (59.9333819 u). The half-life of 2760Co is 2.40 yr. Determine the activity of the radioactive material

Answer 1

Half Life = 2.40 yrs

so lets convert it in seconds

Half Life = 2.40 x 365 days x 24 hrs x 60 min x 60 seconds

=75686400 s

Now

specific activity formula is

activity = (ln(2) / T_1/2 ) x N_A/m ............................ Na is avogadros constant , m = atom mass in g

= (0.6931 / 75686400 s) x ( 6.022 x 10²³ / 59.93g)Bq/g ................... 1Bq=1 decay/second

= 9.20 x 10¹³ Bq/g

Now device contains 1.24 g of cobalt

so activity is = 1.24 x 9.20 x 10¹³ Bq/g = 1.14 x x 10¹⁴ Bq/g