Calculating Half-Life: 50mg to 10mg in 18.4 Days

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    Half-life
In summary, the conversation was about finding the time for a 50-mg sample with a half-life of 18.4 days to decompose to 10 mg. The correct half-life equation is q(t) = q0 2^(-t/tau) and the mistake was made in setting it up as 50(1/2)^n/18.4 = 10(1/2)^n/18.4 instead of 50(1/2)^(n/18.4) = 10. The correct rate was found to be k = -0.0377 and the time it takes to go from 50 to 10 mg was calculated using the equation 10 = 50e
  • #1
runicle
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Something has a half-life of 18.4 days. How much time will a 50-mg sample take to decompose to 10 mg?
My steps
50(1/2)^n/18.4 = 10(1/2)^n/18.4
25^n/18.4 = 5^n/18.4
(5^2)^n/18.4 = 5^n/18.4
5^2n/18.4 = 5^n/18.4
Exponent view
2n/18.4 = n/18.4
2n-n = 18.4/18.4
n=1
Am i doing something wrong?
 
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  • #2
half life equation is
[tex] q(t) = q_{0} 2^{-t/\tau} [/tex]
where q(t) is the quantity after time t,
q0 is the initial quantity
tau is the half life

do u see your mistake?
 
  • #3
Y=CoE^kt
Co = initial amount
Y= what amount is left
K= rate
T= time
First you have to find the rate
Since in 18.4 days you will have 1/2 left you set it up like this:

1/2Co=CoE^kt
Now the Co's cancel and you're left with :
1/2=e^kt
t=18.4
1/2=e^k(18.4)
Take the natural log of both sides now
ln(.5)=lne x 18.4k
-.6931471806 = 1 x 18.4k
divide by 18.4
k = -.0377
Now you must find the time it takes to go from 50 to 10.
10=50e^kt
k we know = -.0377 and we're looking for t

You should be able to figure it out from here.(sorry if my work doesn't look fancy, I'm new to this stuff - I'll pick it up eventually)
 
  • #4
This: 50(1/2)^n/18.4 = 10(1/2)^n/18.4 is your error.
You want to end with 10 g so you want
50(1/2)^(n/18.4)= 10
 
  • #5
Thank you HallsOfIvy for being specific for me:)
 

Related to Calculating Half-Life: 50mg to 10mg in 18.4 Days

1. How do you calculate half-life?

To calculate half-life, you need to know the initial amount of a substance and how much of that substance remains after a certain amount of time has passed. Then, you can use the formula: t1/2 = ln(2)/k, where t1/2 is the half-life, ln is the natural logarithm, and k is the decay constant determined by the specific substance.

2. What is the initial amount of the substance in this problem?

The initial amount of the substance is 50mg, as stated in the problem.

3. How much of the substance will remain after 18.4 days?

Using the given information, we can calculate the decay constant, k, using the formula k = ln(2)/t1/2. Plugging in the values, we get k = ln(2)/18.4 = 0.0377. Then, we can use the formula A = A0e-kt, where A is the final amount, A0 is the initial amount, e is the Euler's number, and t is the time passed. Plugging in the values, we get A = 50e-0.0377(18.4) = 10mg. Therefore, 10mg of the substance will remain after 18.4 days.

4. How many half-lives have passed in 18.4 days?

To determine how many half-lives have passed, we can use the formula Nt/N0 = (1/2)n, where Nt is the final amount, N0 is the initial amount, and n is the number of half-lives. Plugging in the values, we get 10/50 = (1/2)n. Solving for n, we get n = ln(10/50)/ln(1/2) = 2. Therefore, 2 half-lives have passed in 18.4 days.

5. Is half-life a constant for all substances?

No, half-life is not a constant for all substances. It is specific to each substance and can vary greatly. Some substances have a very short half-life, while others have a very long half-life. It all depends on the specific decay constant for that substance.

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