1. If 140Ba is freshly separated from a fission product mixture and found to contain 125 milliCi, (a) what is the activity of 140La that will grow into the sample in 17.2 hr? (b) What is the maximum activity of 140La that will ever be present in the sample, and (c) when will it occur?


Original Answer: Definitely miscalculated this problem and it actually looks simpler than how I attempted below.




Ro = 125 milliCi

T1/2 (Ba) =12.752 days

T1/2 (La) =1.678 days

Decay constants for both 140Ba and 140La


K1 =ln2/ (T 1/2 (Ba)) =0.693/ (12.752 x 24)


K2 =ln2/ (T 1/2 (La)) =0.693/ (1.678 x 24)


b. Calculating the activity of 140La





= 4.13 milliCi













tmax =135.72hours


Corrected Answer/Re-solved: PDF PIC BELOW (REWRITE!)


2. A short-lived radioisotope yttrium-90 can be separated by solvent extraction, leaving the parent radionuclide Sr-90 in the aqueous phase. Their half-lives are T1/2 (Sr-90) = 28.78 y, T1/2(Y-90) = 64 hours=2. 67 days. After splitting phases you have found that the 0.5 mL aliquot of the Sr-eluate counts 250 d/s.


Original Answer: Incorrect


a) How long you can use this Y-free solution of Sr-90 for your research work if the activity of growing Y-90 cannot exceed 10% of Sr-activity?








a. Corrected answer:







b) What will be the total activity of the Sr-eluate after one month?



Corrected/Re-solved Answer: PDF PIC BELOW (REWRITE!)


10 half lives of Y-90 is 30 days without performing calculation


3. A short-lived radioisotope yttrium-90, a medical isotope, is produced by a Sr/Y -90 generator, where the long-lived parent Sr-90 is sorbed on a chromatographic column and yttrium is eluated. The half-lives are T1/2 (Sr-90) = 28.78 y, T1/2(Y-90) = 64 hours=2. 67 days.


a) If the activity concentration of Y-90 in the eluate is 120 kBq/mL, what volume of the physiological solution should be added to obtain a solution with 60 kBq/mL?


Correct Answer: The solution must be diluted two-fold, so the same volume it has now must be added. 60 kBq/mL


b) What time it takes for yttrium to achieve 75% of strontium activity?


Correct Answer: PDF PIC BELOW (REWRITE!)


Original Answer: Incorrect













4. Bones, containing trapped potassium, were found in 1960 in rock with a typical sample yielding a total of 0.25 g K. A fraction (assume 0.000117) of the trapped potassium was 40K, which decays into 40Ar. This sample contained 7.4×10-10 g of 40Ar, of which 9.3×10-11 g was determined to be contaminated from the air. What is the approximate age of the bones?

Original Answer: ehhh, kinda on the right path. Numbers didn’t come out the same…


All potassium = 0.25 g; 40K abundance=0.0117%

40K amount grams = (0.000117) x 0.25 = 0.00002925 g or 2.925 X 10-5 g

Kr moles = (mass of 40 K /atomic mass of 40 K)  = 0.00002925g x 6.022 x 10 23/mol / ( 40g/mol)

= 4.4 x 1017 atoms of 40 K

Ar amount formed = mass of 40 Ar – Ar mass from air

= (7.4 x 10-10 g) – ( 9.3 x 10-11g)  = 6.47  x 10-10 g—–9.74 x 1012 atoms of Ar-40

Needed to solve for the branching ratio PDF PIC BELOW (REWRITE!)


Remaining of original answer…

Ar moles = mass / atomic mass

= (6.47 x 10-10 g) / ( 39.95 g/mol) =  1.6195 x 10-11

1 K40 gives 1 Ar, hence  40K  decayed = Ar formed = 1.6195 x 10-11

40K left = initial 40K – 40K decayed = 7.3125 x 10-7 – (1.6195 x 10-11) = 7.3123 x 10-7

Using the formula

t = (1/k) ln (a/(a-x))

k = decay constant = 5.554 x 10-10 yrs-1

a = initial amount = 7.3125 x 10-

a-x = final amount of 40K left = 7.3123 x 10-7

t = (1/ (5.554 x 10-10 ))   ln ( 7.3125 x 10-7 / 7.3123 x 10-7)

t = 49245 years


5. A measurement of an ore mixture at equilibrium shows 2.1×104 times as many 235U atoms as 231Pa atoms. Calculate the half-life of 235U from the assay data and the known half-life of 231Pa (3.28×104 yr).


Original Answer: Correct!

ʎ0 N0 = ʎp Np


ʎ =ln2/t1/2


(1/ t1/2)0 x N0 = (1/ t1/2)p Np


(1/ t1/2)0 =N0/Np x (1/ t1/2)p

= 2.1×104 x 3.28×104

=6.888 x 108 years


6. In a rock, one finds a nuclidic ratio of 206Pb to 238U of 0.60. What is the age of the rock?


Original Answer: Correct!





7. 222Rn decays to 218Po. At the time t=0, a pure Rn, which under normal pressure and temperature conditions would have volume of 0.65 m3, is captured and sealed in a vial:

a) What time (in min) is needed for accumulation of a maximal amount of 218Po?

b) How much of 218Po will be generated (nanograms)?

c) What will be the ratio of atoms of 222Rn and 218Po when they achieve equilibrium?


Original Answer: Wayyyyy off base….. missing rest of answers….






Corrected Answer/Solved: PDF PIC BELOW (REWRITE!