Question

⁶⁹Cu decays by emitting a β^- particle. This decay process has a half-life of 2.80 min. Answer the following questions about this process and report all answers to three significant figures.

1. Determine the initial activity (in dis/min, do not input units) of a 8.68 mg sample of this isotope. Report your answer to three significant figures in scientific notation.

2. Determine the time (in min) that it takes for the activity of ⁶⁹Cu to decrease to 3.00×10⁶ dis/min.

Answer 1

Solution-
Sol.

(A) As we know that the half life = t1/2 = 2.80 min

So , Decay constant = K

= 0.693 / t1/2 = 0.693 / 2.80 = 0.2475 min-1

So the , Mass of Cu = 8.68 mg = 8.68 * 10^-3 g

And the molar Mass of Cu = 69 g/mol

Now the, initial number of nuclei present

= ( Mass of Cu / Molar Mass of Cu ) × Avogadro's no.

= ( ( 8.68 * 10^-3 ) / 69 ) * 6.022 * 10^23

= 7.57 * 10^19

Hence the , initial activity of Cu = A°

= initial number of nuclei present * Decay constant

= (7.57 *10^19)* 0.2475

= 1.87 * 10^19 dis/min

(B)
A = activity of Cu after time t =3.00*10^6 dis/min

Therefore , ln ( A° / A ) = K t

ln((1.87 * 10^19 )/(3.00*10^6)) = 0.2475 t

0.2475 t = 29.46

t = 119.03 min