Question

Cobalt-60, which undergoes beta decay, has a half-life of 5.26 yr.

Part A

How many beta particles are emitted in 250s by a 2.90mg sample of 60-Co?

Part B

What is the activity of the sample in Bq?

Answer 1

Part B:

t_1/2 = 0.693/⋋, where t_1/2 is the half life of radioactive nuclide and ⋋ is its decay constant.

5.26 yr = 0.693/⋋, i.e. ⋋ = 0.132/yr. = 4.18*10^-9/s.

Activity = ⋋*(x/A)*N_A, where N_A = 6.023*10²³., x = weight of sample, i.e. 2.9*10^-3 g, A = mass no. of ⁶⁰Co, i.e. 60.

Activity = 4.18*10^-9 * (2.9*10^-3 / 60) * 6.023*10²³.

= 1.217*10¹¹ Bq or dps (disintegration per sec).

Part A:

No. of beta particles emitted in 250 s = 250 * Activity

= 250 * 1.217*10¹¹

   = 3.04*10¹³.