SOLUTION: The half life of a certain isotope is 8.5 hours. If we have a sample of 20 ounces of that isotope, how much of it will remain after one full day? The correct answer is one of th

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: The half life of a certain isotope is 8.5 hours. If we have a sample of 20 ounces of that isotope, how much of it will remain after one full day? The correct answer is one of th      Log On


   

Question 1199030: The half life of a certain isotope is 8.5 hours. If we have a sample of 20 ounces of that isotope, how much of it will remain after one full day?
The correct answer is one of the following. Which one is correct?
A) 0.28 ounces
B) 0.14 ounces
C) 1.41 ounces
D) 2.82 ounces
Found 3 solutions by ikleyn, MathTherapy, math_tutor2020:
Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.

One full day is 24 hours.


24 hours is  how many half-lives ? - It is  24%2F8.5 = 2.82 half-lives, i.e. close to 3 half-lives.


Hence, after 24 hours, the remaining amount will be  about  1%2F2%5E3 = 1%2F8  of the initial amount, which is 20 ounces.


So, we can expect  something about 20%2F8 = 2.5 ounces for remaining mass.


The closest option is D).    ANSWER


All other optional masses are much lesser and much further in time than 3 half-lives; 
so, this MENTAL estimation is just enough to make a final conclusion.

Solved.



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
The half life of a certain isotope is 8.5 hours. If we have a sample of 20 ounces of that isotope, how much of it will remain after one full day?
The correct answer is one of the following. Which one is correct?
A) 0.28 ounces
B) 0.14 ounces
C) 1.41 ounces
D) 2.82 ounces 
If 1%2F2-life consists "a" time periods, the DECAY CONSTANT, or matrix%281%2C3%2C+k%2C+%22=%22%2C+ln%281%2F2%29%2Fa%29

                                                So, we have: matrix%281%2C3%2C+k%2C+%22=%22%2C+ln%281%2F2%29%2F8.5%29                                         
                                                             

                            OONTINUOUS GROWTH/DECAY formula: matrix%281%2C3%2C+A%2C+%22=%22%2C+A%5Bo%5De%5E%28kt%29%29
                                                 So, we get: matrix%281%2C3%2C+A%2C+%22=%22%2C20e%5E%28-+.0816%2824%29%29%29 --- Substituting  

Amount remaining after 1 full day (24 hours): A = 2.82168 = 2.82 (ROUNDED to 2 decimal places) ounces (CHOICE D).

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Half-life Formula
y = a*(0.5)^(x/H)

We have:
a = 20 = starting amount in ounces
H = 8.5 = half-life in hours
x = 24 = number of hours
y = number of ounces left after x hours

So,
y = a*(0.5)^(x/H)
y = 20*(0.5)^(24/8.5)
y = 2.82528945857332
y = 2.83
After a full 24 hours, there should be about 2.83 ounces of material remaining.

It appears there might be some rounding error either on my part, or perhaps your teacher made a typo somewhere.

The closest value to this is D) 2.82 ounces which is likely the final answer.