Question

1.) An old vial of radioactivee material was found buried in the ground at Hanford Site. in 2014, it was found to contain 0.1578 Ci of 238Pu and 0.1125 Ci of 234U. the vial was labebled year 1951. how many atoms of 238 Pu and 234U were present in the vial in 1951 ?

238 94U a (t1/2)=87.72 y -> 23492U a (t1/2)=2.455*10^5 y ->23090TTh

Answer 1

Calculation for Pu²³⁸:

Half life period (t^1/2) of PU ²³⁸= 87.72*31536000 = 2766337920 sec

Available quantity of PU²³⁸ = 0.1578 Ci

(t^1/2) = 0.693/k

k = 0.693/(t^1/2)

= 0.693/2766337920

= 2.5*10^-10

From the equation:

K = (2.303/t)*log(Initial concentratioin)/(final concentration)

2.5*10^-10 = (2.303)/ (2014-1951 time period) log (Initial concentratioin/0.1578

2.5*10^-10 = (2.303)/ (63) log (Initial concentratioin)/0.1578

(2.5*10^-10)/0.003655 = log (Initial concentratioin)/0.1578

6.83*10^-8 = log (Initial concentratioin)/0.1578

Antilog (6.83*10^-8 ) = (Initial concentratioin)/0.1578

1.0000001 = (Initial concentratioin)/0.1578

Initial concentratioin = (0.1578)*(1.0000001)

Initial concentratioin of Pu ²³⁸ in 1951 = (0.1578) *3.7*10¹⁰ decays per sec = 0.58386*10¹⁰ dps.

Calculation for U²³⁴:

Half life period (t^1/2) of U ²³⁴= 2.455*10⁵*31536*10³ = 77420.88*10⁸ sec

Available quantity of U²³⁴ = 0.1125 Ci

(t^1/2) = 0.693/k

k = 0.693/(t^1/2)

= 0.693/77420.88*10⁸

= 8.95*10^-14

From the equation:

K = (2.303/t)*log(Initial concentratioin)/(final concentration)

8.95*10^-14 = (2.303)/ (2014-1951 time period) log (Initial concentratioin/0.1125

8.95*10^-14 = (2.303)/ (63) log (Initial concentratioin)/0.1125

(8.95*10^-14)/0.003655 = log (Initial concentratioin)/0.1125

2.448*10^-11 = log (Initial concentratioin)/0.1125

Antilog (2.448*10^-11) = (Initial concentratioin)/0.1125

1.00000000006 = (Initial concentratioin)/0.1125

Initial concentratioin = (0.1125)*(1.00000000006)

Initial concentratioin of U ²³⁴ in 1951 = (0.1125) *3.7*10¹⁰ decays per sec = 0.416*10¹⁰ dps.