The Curie (Ci) is a unit of radioactivity named after Marie Sklodowski-Curie and

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 T1/2 1600 years. Thus the decay coefficient for 226Ra equals k = (ln 2)/T1/2 4. 33 10-4 1/year

Fill in the blanks.

a. As the time elapses from t =0 by the step t = 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 t = 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%

Explanation / Answer

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%

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