Differential Equations and Radioactive Decay

[Northbysouth]

Homework Statement

The radioactive decay of a substance is proportional to the present amount of substance at any time t. If there was 15 grams at t=0 hours and 10 grams at t=3 hours. Set up the differential equation that models this decay and use the method of separation of variables to solve for the equation that will give the amount of the substance at any time t. Find when the half-life occurs and the amount of substance at t=10 hours.

Homework Equations

The Attempt at a Solution

So I have the equation

N = De^kt

for the decay

Plugging in N = 15 grams at t = 0 gives mt

D = 15

Then plugging in t=3 and N = 10 grams

10 = 15e^3k

I get k = ln(10)/45

The half life occurs at 0.5N

0.5N = 15e^{ln(10)/45 * t}

Solving for t I get

t = (45*7.5)/300

t = 1.125 hours

But this doesn't make sense. Was I wrong to assume that N = 7.5 at the half-life time?

 

[Ray Vickson]

Homework Statement

The radioactive decay of a substance is proportional to the present amount of substance at any time t. If there was 15 grams at t=0 hours and 10 grams at t=3 hours. Set up the differential equation that models this decay and use the method of separation of variables to solve for the equation that will give the amount of the substance at any time t. Find when the half-life occurs and the amount of substance at t=10 hours.

Homework Equations

The Attempt at a Solution

So I have the equation

N = De^kt

for the decay

Plugging in N = 15 grams at t = 0 gives mt

D = 15

Then plugging in t=3 and N = 10 grams

10 = 15e^3k

I get k = ln(10)/45

The half life occurs at 0.5N

0.5N = 15e^{ln(10)/45 * t}

Solving for t I get

t = (45*7.5)/300

t = 1.125 hours

But this doesn't make sense. Was I wrong to assume that N = 7.5 at the half-life time?

How did you go from 10 = 15e^3k to k = ln(10)/45 ? (It is wrong, BTW.)

 

[Northbysouth]

Yes, I think I made a mistake there.

I think it should be:

k = ln(10/15)t/3

When I use this, plugging in N =7.5 for the half-life N value, I get t = 2.25 which doesn't seem right either.

 

[Ray Vickson]

Yes, I think I made a mistake there.

I think it should be:

k = ln(10/15)t/3

When I use this, plugging in N =7.5 for the half-life N value, I get t = 2.25 which doesn't seem right either.

You need to review material on logarithms. How do you get log(10/15)?

 

[HallsofIvy]

Homework Statement

The radioactive decay of a substance is proportional to the present amount of substance at any time t. If there was 15 grams at t=0 hours and 10 grams at t=3 hours. Set up the differential equation that models this decay and use the method of separation of variables to solve for the equation that will give the amount of the substance at any time t. Find when the half-life occurs and the amount of substance at t=10 hours.

Homework Equations

The Attempt at a Solution

So I have the equation

N = De^kt

HOW did yo get this? There was no "k" in the problem. Did you, as the problem requires, actually set up the diffrential equation and solve?

for the decay

Plugging in N = 15 grams at t = 0 gives mt

D = 15

Then plugging in t=3 and N = 10 grams

10 = 15e^3k

I get k = ln(10)/45

The half life occurs at 0.5N

0.5N = 15e^{ln(10)/45 * t}

Solving for t I get

t = (45*7.5)/300

t = 1.125 hours

But this doesn't make sense. Was I wrong to assume that N = 7.5 at the half-life time?

 

[Ray Vickson]

In my second reply, where I asked "how do you get ln(10/15)", I meant "how do you calculate ln(10/15)"? The ln(10/15) part is OK, but somewhere after that you made an error.

(I wanted to put this into an edited version of my response, but for some reason the "edit" option is now unavailable.)

 

[Northbysouth]

I think I misinterpreted the question. My professor did a similar example in class and I assumed that I could use the equation that she used.

So, unless I'm much mistaken, I believe that my initial equation is:

dN/dt = kt

dN = kt dt

∫dN = ∫kt dt

N = kt²/2 + C

When t=0, N =15

C = 15

N = kt²/2 + 15

At t=3, N=10

10 = k(3)²/2 + 15

k = -10/9

N = -10t²/18 + 15

The half-life should occur when N=7.5

7.5 = -10t²/18 + 15

t = 3.674

However, when I go to check the amount of the substance left at t=10,

N = -10(10)²/18 + 15

N=-40.56

I take this to mean that N is actually equal to 0, unless I've made a mistake in my work. Am I making sense here?

 

[Ray Vickson]

I think I misinterpreted the question. My professor did a similar example in class and I assumed that I could use the equation that she used.

So, unless I'm much mistaken, I believe that my initial equation is:

dN/dt = kt

dN = kt dt

∫dN = ∫kt dt

N = kt²/2 + C

When t=0, N =15

C = 15

N = kt²/2 + 15

At t=3, N=10

10 = k(3)²/2 + 15

k = -10/9

N = -10t²/18 + 15

The half-life should occur when N=7.5

7.5 = -10t²/18 + 15

t = 3.674

However, when I go to check the amount of the substance left at t=10,

N = -10(10)²/18 + 15

N=-40.56

I take this to mean that N is actually equal to 0, unless I've made a mistake in my work. Am I making sense here?

You are mistaken: your DE dN/dt = kt says that the rate of decay is proportional to time, but the question said it should be proportional to N.