Question

The Curie (Ci) is a unit of radioactivity named after Marie Sklodowski-Curie and Pier Curie, it represents the activity of one gram of the most stable isotope of radium ²²⁶Ra, or 3. 7 ∗ 10¹⁰ decays per second. While Ci is not a SI unit, it is commonly used by the US industries and government bodies. The isotope has a half-life of about T_1/2 ≈ 1600 years. Thus the decay coefficient for ²²⁶Ra equals k = (ln 2)/T_1/2 ≈ 4. 33 ∗ 10^-4 1/year

Fill in the blanks.

a. As the time elapses from t =0 by the step Δt = 1/k ≈ 2308 years, the percentage of an original radioactive sample of ²²⁶Ra remaining drops by the factor of e___? and equals ___________ %.

b. As the time elapses from t =0 by the three step 3 ∗ Δt = 3/k ≈ 6924 years, the percentage of ²²⁶Ra remaining drops by the factor of e___? and equals ___________ % of the original amount.

P(t)/P(0) ∗ 100% = e^-kt∗ 100%

Answer 1

a.

P(t) = P(0)* e^-kt

At, t = 0, P(t) = P(0) = P(0)* e^-k*0 = P(0)* e⁰

At, t = 1/k, P(t) = P(0)* e^-k*(1/k) = P(0)* e^-1

So, the percentage of an original radioactive sample of ²²⁶Ra remaining drops by the factor of

= [P(0)* e⁰]/[ P(0)* e^-1] = e¹

%remaining = P(t)/P(0) ∗ 100% = e^-kt∗ 100% = e^-1∗ 100% = 36.79%

b.

At, t = 0, P(t) = P(0) = P(0)* e^-k*0 = P(0)* e⁰

At, t = 3/k, P(t) = P(0)* e^-k*(3/k) = P(0)* e^-3

So, the percentage of an original radioactive sample of ²²⁶Ra remaining drops by the factor of

= [P(0)* e⁰]/[ P(0)* e^-3] = e³

%remaining = P(t)/P(0) ∗ 100% = e^-kt∗ 100% = e^-3 * 100% = 4.98%