ln 2 ln 2 t 1 = = = 26.06days λ.0266 = = 37.6days

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Colin Williams
5 years ago
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1 Pae 1 1/3/ Problem Set 3 Homework Solutions 1. (3 points The activity of a radioisotope is found to decrease to 45% of its oriinal value in 3 days. (a What is the decay constant? We are solvin for the decay constant, λ. We know that the activity after t3 days is now.45a of the oriinal activity, A. Solvin for λ: A A e λt A.45 e λ (3 (b What is the half-life? By definition of half-life: (c What is the mean life? By definition of mean life: A λ.266days min 1 ln 2 ln 2 t days λ τ days λ (4 points How lon will it take for each of the followin radioisotopes to decrease to.1% of its initial activity? We want to solve for time usin the equation: A A λt e where A/A is.1 and λ is specific to each of the atoms. (a 64 Copper, t ½ 12.7 hr, so λ.55 hr -1, t hr (b (c (d 41 Scandium, t ½ ms, so λ.12 ms -1, t 1.19x1 4 ms 99 Technetium, t ½ 2.12x1 5 y, so λ 3.27x1-6 y, t 4.23x1 6 y 99m Technetium, t ½ 6.2 h, so λ.115 h, t 12 h

2 Pae 2 1/3/26 3. (4 points A sample contains 1. GBq of 9 Sr and.62 GBq of 9. The decay equation for 9 Sr: 9 t 1 / y 9 t 1 / 2 64h 9 38 Sr 39 4 Zr(stable λ Sr 7.55x1-1 s -1 and λ 3.8x1-6 s -1 A,Sr 1.x1 9 s -1 and A,.62x1 9 s -1 9 Sr and 9 are in secular equilibrium. (a What will be the total activity of the sample 1 days later? Usin the eneral case for serial radioactive decay but modifyin it for an initial amount of dauhter activity: N λ Sr N Sr λ λ Sr (e λ Srt e λ t + N e λ t Multiplyin the equation by λ allows for the equation to be written in activities: A λ A Sr (e λ Sr t e λ t + A λ t e λ λ Sr 3.8x1 6 1x1 9 6 (e 7.55x e 3.8x x x x1 9 e 3.8x GBq A A e λt Sr Sr 1x1 9 e 7.55x GBq A total GBq 1.97GBq (b What will be the total activity of the sample years later?

3 Pae 3 1/3/26 A λ A Sr (e λ Sr t e λ t + A e λ t λ λ Sr 3.8x1 6 1x (e 7.55x e 3.8x x x x1 9 e 3.8x GBq A A e λt Sr Sr 1x1 9 e 7.55x GBq A total.5 +.5GBq 1GBq 4. (2 points The lobal inventory of 14 C is about 8.5 x 1 18 Bq and we are told that all the activity comes from cosmic rays interaction. The activity tells us the number of atoms decayin per second. We can convert that to the mass of 14 C that is decayin per second and multiply by a year to determine the mass per year that is decayin and needs to be replenished to keep the activity steady: 18 atoms 1mol x k 23 sec 6.23x1 atoms 1mol 5. (4 points Calculate the number of rams contained in a 5 mci source of the followin nuclides. For each atom, we will calculate the specific activity and then use it to find the weiht of each atom: 6.2x1 23 λ 4.17x1 23 SA M M t 1/ 2 (a 18 F, t 1/ min SA 4.17x x1 18 Bq 1 9.5x1 7 Ci

4 Pae 4 1/3/26 # rams 5x1 3 Ci 5.25x x1 Ci (b 14 C, t 1/2 573 y 4.17x SA 1.65x1 Bq 4.45Ci # rams 5x1 3 Ci Ci (c 32 P, t 1/ days 4.17x SA 1.55x1 Bq 2.85x1 Ci # rams 5x1 3 Ci 1.75x x1 5 Ci (d 235 U, t 1/2 7.38x1 8 y 4.17x SA 6 8x1 Bq 2.16x1 Ci x # rams 5x1 3 Ci x1 4 Ci The idea is that even thouh all four atoms have the same number of radioactive nuclei, they all weih differently. 6. (3 points Charcoal found in a deep layer of sediment in a cave is found to have an atomic 14 C/ 12 C ratio only 3% that of a sample from a hiher level with a known ae of 185 years. What is the ae of the deeper layer? 12 C is stable, but 14 C decays with a half-life of 573 yrs. The exponential decay law is: N N e λt In this case we are iven a ratio of the number of 14 C to 12 C, but since 12 C is stable the number of atoms will not chane, only the number of atoms of 14 C will. Therefore we can write the equation as:

5 Pae 5 1/3/26 N 14 N C 14 C e N N C C We assume that the upper and lower layer of the sediment will have the same initial amount of 14 C to 12 C ratio upon formation. Calculatin then the ratio of the number of atoms for the upper sediment: λ 14 C t N ln C e N 12 C We are told that the lower level has a 3% of the upper level atomic ratio, therefore (.8*(.3.24 of the ratio of the oriinal amount upon formation. Calculatin now for time: ln 2 t.24 e 573 t 1182 y 7. ( 5 points Technicium 99m ( 99m Tc is used for nuclear medicine imain procedures. The minimum dose that enerates a useful imae is 5 mci. Generally, each patient receives 1-15 mci for an imain procedure. 99m Tc is produced by elution from enerators containin the bound parent isotope 99 Mo. A radioisotope enerator is calibrated to contain 5 mci of 99 Mo at 12 noon on Friday, Sept 22. The enerator containin 99 Mo is delivered on Monday mornin at 6 am. (a The 99m Tc is eluted at 8 am on Monday, Sept 25. How much 99m Tc is eluted (in mci? This is the decay chain: 99 t 1 / h 99m t 1 / 2 6.2h 99 Mo Tc Tc Lets use the eneral case to solve for the amount of 99m Tc on Monday at 8 AM, a total of 68 hours after the enerator is calibrated. The initial activity amount of 99m Tc is zero. Also, λ Mo 2.92x1-6 s -1 and λ 99mTc 3.2x1-5 s -1. Also the activity converted to Bq: 1.85x1 1 s -1. Solvin:

6 Pae 6 1/3/26 λ 99 A A 99 m Tc Mo (e λ Mo t e λ 99 m t Tc m Tc λ λ 99 m Tc Mo 3.2x x1 1 (e 2.92x e 3.2 x x x x1 9 Bq 267mCi (b The enerator is then eluted on Wed, Sept 27 at 8 am. Is there enouh 99m Tc available for 1 patient imain studies? First we need to calculate how much Mo is now left at this time, 68 hours later: A A e λt 5mCi e 2.92 x mCi 9.5x1 9 Bq This is now our new A for the Mo. Aain usin the eneral case, but now the time that has passed is 48 hours: A 99 m Tc λ 99 m A Tc Mo λ t (e λ Mo t e λ λ99 m Tc Mo 99 m Tc 3.2x x1 9 (e 2.92x e 3.2 x x x x1 9 Bq 161mCi This will just be enouh for 1 patients to receive 15 mci. (c What is the last date that the enerator can be used to produce enouh 99m Tc for a patient study (i.e., > 5 mci? We need to find the time t when the activity of 99m Tc is 5mCi. Since we are in transient equilibrium we can use equation 4.42 to find the activity of the 99 Mo when the activity of 99m Tc is 5 mci: λ 99 A A m Tc Mo 99 m Tc λ λ 99 m Tc Mo 3.2x1 5 A 1.85x1 8 Mo 3.2x x1 6 A Mo 1.68x1 8 Bq Now solvin for time at which this is the activity of the parent: λt A Mo A Mo e 1.68x x1 1 e 2.92x1 6 t t 18.6days

7 Pae 7 1/3/26 8. ( 2 points The key to this problem is to realize that the parent, 238 U is in secular equilibrium with its dauhter, 226 Ra. With secular equilibrium, the activity of the parent and dauhter is equal: λ U N U λ Ra N Ra λ U λ Ra N Ra ln x x1 1 y N U 162 t 1/ x1 9 y Appendix D lists the half life for 238 U as 4.468x1 9 y. Our estimation is 17% off, pretty close usin such a simple methodoloy! 9. ( 3 points. Elemental analyses on 5 separate samples from the same meteorite ave the followin data, expressed in atoms/sample. How old is this meteorite? Sample 87 Rb 87 Sr 86 Sr Plottin the data: y.632x D/D' P/D' 1 y 1 9 t ln 1+ ln( x1 y λ x ln x1 1 y

Recap from last time

Recap from last time Nuclear Decay Occurs. when a nucleus is unstable (lower open energy levels) An unstable nucleus metamorphoses ( decays ) into a more stable (more tightly bound) nucleus Difference