Problem

The half-life of cesium-137 is 30 years. Suppose …

05:33

Problem 8 Medium Difficulty

Strontium-90 has a half-life of 28 days.
(a) A sample has a mass of 50 $\mathrm{mg}$ initially. Find a formula
for the mass remaining after $t$ days.
(b) Find the mass remaining after 40 days.
(c) How long does it take the sample to decay to a mass
of 2 $\mathrm{mg} ?$
(d) Sketch the graph of the mass function.

Answer

a) $$
y(t)=50 e^{-0.0248 t}
$$
b) $$
18.5 \mathrm{mg}
$$
c)$$ 129.8~days$$
d) See video for graph

Related Courses

Calculus 1 / AB

Essential Calculus Early Transcendentals

Chapter 3

INVERSE FUNCTIONS

Section 4

Exponential Growth and Decay

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Video Transcript

Hey, guys, welcome back This problem We know the initial value is 50 milligrams that there is 1/2 life of 28 days when it says that there is 1/2 life of 28 days. That means that when t z 28 basically one why 28? That will be half of the initial or half of 50 which will be 25. Because it's an exponential growth problem. We know that it will follow the form changing. Why Changing t Okay, why? And if you integrate that, that comes out to why so two y o e raised to the Katie. So now what we want to dio is gonna solve for K plugging in our known values. We know that when t is 28 that why will be 25. So we have 25 Cols Weiss abo We know that why Sebo is 50. The initial e raised to the K is unknown and we're ready Said that tea was 28 I never want to solve for K. We divide both sides by 50 we get 1/2 e to the 28 k Ln 1/2 Take the Elena Both sides 28 times k k is equal to the natural log of 1/2 divided by 28. That comes out negative 0.0 to for eight. Now we can plug back in to this equation with their known values. You get that? Why solitude? Why subzero, which is 50 times e negative 0.2 four a. T. That is our equation for why, based on our time T now we're asked to figure out how much remains after 40 days Part B appear What happens when TZ little 40 so is plugged into that equation, which is found y 0 50 e race native 0.248 applied by 40. And when we do the math there, we get a value of 18.5. So we know that why of 40 it's cool to 18.5 milligrams. Next one. If you're on how much time it takes for the sample to reach two milligrams, want to say to milligrams? Look into that same equation. But now we're solving for teeth 0 to 50 e to the negative 0.248 Blood by t divide both sides by 50. I knew that you get 0.4 seems to eat native 0.2 for 80. The natural log on both sides You get Ellen 0.4 seal to negative 0.24 80 He didn t is the natural log of 0.4 divided by negative 0.0 248 And when you do that, your left with 100 and 29.8 So t 0 to 129 0.8 days, that's how long it will take. Take it down to just two milligrams and now for part deeds. As to graft the function, I went ahead. I grafted using Dismas just down here, where we have some more broom. Few things you could note about this graph. You can see that right here. Sometimes you go to zero. The initial values 50. Like we said before it appeared. The initial value is 50 and that'll slowly approach zero as time goes on. This helped Thanks for watching

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