1) You are in charge of an antibody serum that has been ordered for treatment of a patient with a serious gastrointestinal problem. If the antibiotic serum has passed the shelf-life, it will not be as effective. The best application date, as advertized on the label, is not to exceed nine days from receipt of serum, and you are to administer the serum on June 13. Will the product remain viable on that date if it was received on the 4th of June? The l00 gram sample of microbiota 247 Tb you ordered, has a half-life of 2.70 days, and you are to administer exactly 13.25 grams of the material to the patient. Will you have enough material for administration?

2) You have to inventory various microbiological antigen serum samples in the hospital laboratory, and you observed that there is a 50.0 gram sample of antigen T-cell serum that has denatured to 12.5 grams of viable macrobiotic activity in just 14.5 days on the self. What is the antigen T-cell serum half-life in your opinion?

3) You are to reorder microorganism activation products so that the active ingredient of the microbe is always viable. There is a microbiological product (Ag-235) that has a half-life of 12.4 hours. How much of a 750 gram sample of active ingredient would be left in the solution container after 62.0 hours of exposure to the laboratory environment?

4)You are to reorder the original volume of Ab-325, but the invoice and order information has been lost somehow, and you do not want to contact the manufacturer and appear stupid, so you calculate the original volume of Ab-325 by evaluating what is left of the original order. You weigh the Ab-325 and determine that the volume remaining in the container after 40.35 days is exactly 5.0 grams. The label on the bottle of Ag-325 informs you that the half-life of the active ingredient is 8.07 days. As a result of your calculations, you determine that the original order volume last month was exactly X-X grams of Ab-325.

5) As a nurse practitioner you are to administer exactly 100 grams of CT-cell globulin ointment to a patient's surgical scares to prevent serious infection. The half-life of the ointment is preciously 1.4 X 1024 nanoseconds upon opening the solution container, so application to the patient's scaring must be extremely fast. If you have 25.0 grams of CT-cell ointment left after 2.8 X 1024 nanoseconds of application time to the patient, how many grams were in the original sample of CT-cell ointment?

6) You noticed that a five hundred gram sample of PP-tt1 radio-isotope for X –ray film treatment has decayed to sixty-two point five grams in just six hundred thirty-nine thousand seconds. You have to determine the half-life of the radio-isotope PP-tt1 in order to maintain proper X-ray film processing volumes for your laboratory use. Therefore, what is the half-life of this radio-isotope? Your boss wants to know!

1. The first part is simple arithmetic.

For the second part, you'll have to use:

N = N_o \(\frac12\)^{t/h}

where N is the number of grams you want to calculate,
No is the initial number of grams,
t is the number of days since the reception date, until the date it is used,
h is the half life.

If you get N \geq 13.25 g, then it's enough.

2. Use that equation again.

N = N_o \(\frac12\)^{t/h}

Here, N = 12.5 g, No = 50.0 g, t = 14.5 days, h = ? Days.


Can you give the others a try? (and post what you did)

OK... sorry I'm so lost with this, how do u calculate that on log calculator?

I'm not sure what you are asking now...

For 1. type in:
100
X
0.5
^ [this is caret, or power, often given by the button x^y]
(
9
\div
2.7
)
=

That's what you type on a scientific calculator.

Otherwise...

9 \div2.7
M+
0.5 ^ MR
X 100

O OK... so then it would be 9.92 grams so there wouldn't be enough since you need 13.25 grams. Right?

Yes, that's correct :)

1) Yes, the product will be viable because it is barely 9 days from June 4 to June 13.
100g x 5^ (9 days/2.70 days) =10 grams
No, there isn't enough material for administration because we need 13.25 grams.

2) 7.25 days (2 half-lives)
14.5 x log 2 / log (50.0g/12.5g) =7.25 days
50.0/2= 25
25/2= 12.5
It takes 2 half-lives to undergo decay from 50 to 12.5.

3) 62 hours/12.4 hours=5 (undergoes 5 half-lives)
750/2=375
375/2=187.5
187.5/2=93.75
93.75/2=46.8
46.8/2=23.4 grams
There will be 23.4g of Ag-235 in the sample after 62.0 hours.

4) 40.35 days/ 8.07 days= 5 (undergoes 5 half-lives)
5.0/2=2.5
2.5/2=1.25
1.25/2= .625
.625/2= .3125
.3125/2= .15625 x 100= 15 grams
There was 15 grams in the original order.

5) 2.8 x 10^24/1.4 x 10^24=2 (undergoes 2 half-lives)
25.0 x 2= 50
50 x 2=100 grams
There was 100g in the original sample of the CT-cell ointment.

6) 500/2= 250
250/2= 125
125/2= 62.5
It takes 3 half-lives to undergo decay from 500 grams to 62.5 grams.

Thanks so much you gave me an idea on how to solve these problems (=
and I got an A on my assignment!! =D
Thanks Again

4. That's not what I told you to do... nor is what the question asking for :confused:

5 g is what remains AFTER the decay. The question asks for the original mass.

6. Yes, that takes 3 half lives, but the actual question is "What is the half life of the isotope?"

You are to find the duration of the half life.

3 half lives take 639000 s
1 half life takes ? s

Otherwise, well done on the others! :)