exponential decay: Bismuth-210 has a half-life of 5.0 days.

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Phenomniverse

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My math tutor used the following example to demonstrate the idea of exponential decay:

Bismuth-210 has a half-life of 5.0 days.
a) Suppose a sample originally has a mass of 800mg. Find a formula for the mass remaining after t days.
b) Find the mass remaining after 30 days.
c) When is the mass reduced to 1mg?

He solved it using the formula P = P_0e^{-kt} where P_0 is the initial mass (800mg), t = 5 days and P = 400 (because we know that the mass will have reduced by half after 5 days) and then solving for k to work out what the decay constant is.

I proposed an easier solution, based on the idea that repetitive halving is the same as dividing by two to the power of n, where n is the number of times to halve the initial number. In this case, that will be t/5. So my formula was M_t=M_0/[2^(t/5)], where M_t represents the mass at time t, and M_0 represents the initial mass. My solution seems to work, but when it comes to part (c) from the example, I'm stuck as to how to solve for t.
bismuth q.png

How can I get that t on its own?
 
\(\displaystyle 2^{t/5} = 800\)

\(\displaystyle (2^{t/5})^5 = (800)^5\)

Continue …
 
Phenomniverse said:
My math tutor used the following example to demonstrate the idea of exponential decay:

Bismuth-210 has a half-life of 5.0 days.
a) Suppose a sample originally has a mass of 800mg. Find a formula for the mass remaining after t days.
b) Find the mass remaining after 30 days.
c) When is the mass reduced to 1mg?

He solved it using the formula P = P_0e^{-kt} where P_0 is the initial mass (800mg), t = 5 days and P = 400 (because we know that the mass will have reduced by half after 5 days) and then solving for k to work out what the decay constant is.

I proposed an easier solution, based on the idea that repetitive halving is the same as dividing by two to the power of n, where n is the number of times to halve the initial number. In this case, that will be t/5. So my formula was M_t=M_0/[2^(t/5)], where M_t represents the mass at time t, and M_0 represents the initial mass. My solution seems to work, but when it comes to part (c) from the example, I'm stuck as to how to solve for t.
View attachment 10027

How can I get that t on its own?
Click to expand...

I would start by taking the log (base 10 or base e) of both sides, and simplifying.
 
I went with this (see attached pic)
bismuth part c.jpg

I got to the same answer as my tutor did with his method, so I assume its legitimate..?
 
How did you evaluate log2(800) ?

The instructor might like to see your final step. :cool:
 
mmm4444bot said:
How did you evaluate log2(800) ?

The instructor might like to see your final step. :cool:
Click to expand...

I did it with my calculator. My calculator doesn't have a log base n button, so I followed some youtube intstructions which said to do log_10(800)/log_10(2).
 
mmm4444bot said:
I'm not sure I follow this. It seems like you've written that the mass is 1 mg, at some time t.
Click to expand...


Yes, that's right. The mass degenerates over time, so the question is how many days does it take for the original 800mg to degenerate to 1mg.
 
Phenomniverse said:
… log_10(800)/log_10(2)
Click to expand...
That's called the change-of-base formula. This is what the instructor might want to see, as your last step, before reporting the decimal approximation for t.

(Unless your instructor is a machine, heh.)

Good job.
 
Phenomniverse said:
… the question is how many days does it take for the original 800mg to degenerate to 1mg.
Click to expand...
Thanks. I need to clean my glasses! :oops:

My steps were similar.

2^(t/5) = 800

2^t = 800^5

ln(2^t) = ln(800^5)

t∙ln(2) = 5∙ln(800)

t = 5∙ln(800)/ln(2)
 
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