Find the amount of phosphorus-32 remaining after 84 days.

Question

Phosphorus-32 is used to study  plants use of fertilizer \, It has a half-life of 14.3 days. write the exponential decay function of  a 50-mg sample. Find the amount of phosphorus-32 remaining after 84 days 

Answer 1

Extending Ann's answer, there are several ways to write the decay equation to compute the remaining mass (M) from the initial mass (M₀). 

 

One of them uses the power of 1/2 and the number of half-lives (t/t_½):

 

t_½=14.3 days

 

M =M₀ (1/2)^t/t_½= M₀ 2^-t/t_½

 

 

Another way is to use the e, with the decay constant, λ = ln(2)/t_½:

 

M = M₀ e^-λt

 

 

Both of them will give the same amount of P-32 after 84 days.

 

(50 mg)2^-84/14.3 = 0.85 mg

 

(50 mg)e^{-(84)ln(2)/14.3} = 0.85 mg

Answer 2

 

 

Every 14.3 days, 1/2 of the mass of Phosphorus-32 will be lost.  Beginning with 50 mg. :

 

After      14.3 days                 - 25.00 mg. remaining

After 2 (14.3) or 28.6 days      -12..50 mg. remaining

After 3 (14.3) or 42.9 days     -   6.25 mg. remaining

After 4 (14,3) or 57.2 days      -  3.125 mg remaining

After 5 (14.3) or 71.5 days      -  1.5625 mg. remaining

After 6(14.3) or  85.8 days      -   0.78125 mg. remaining

 

So your answer will lie somewhere between 1.5625 mg. (after 71.5 days) and 0.78125 mg. (after 85.8 days).

 

Every 14.3 days, you lose 1/2 or the original mass of Phosphorus-32.

 

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The answer I found below, on Google, seems to agree with what I have (above).  0.85 mg. falls between 71.5 days and 85.8 days.

 

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84 / 14.3 = 5.87412587 finding the number of times half life is passed.

to find how much is left will be one over 2 raised to the 5.874 power

or 1/58.64 remaining

0.017 or 1.7 percent of 50 mg.

1.7 percent of 50 mg is 0.85 mg.
You might want to check my math, I didn't have a calculator or the function before I started.

exponential decay function
y=a(1-r)^x
a = initial amount
r = growth rate
x = time interal
http://www.regentsprep.org/regents/math/...