# The rate of decay of particular isotope of Radium (in mg per century) is proportional to its mass (in mg). A 50mg sample takes one century to decay to 48mg. Ho0w long will it take before there are 45 mg of the sample?

Amount of Radium after

It will take 2.6 centuries for the Radium to weigh 45mg.

Let us define the following variables:

Then

Which we can integrate to get:

And so the Specific Solution is:

[ We should just check that we have not made a mistake by checking the initial condition:

so we know the solution is sound]

Hence it will take 2.6 centuries.

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We can model the rate of decay using the differential equation:

dm/dt = -k * m

where m is the mass of the isotope at time t, and k is the decay constant.

Given that the rate of decay is proportional to the mass, we can write:

dm/dt = km

Using the initial condition that a 50 mg sample decays to 48 mg in one century:

dm/dt = k * 50, when m = 50

So, k = (dm/dt) / 50 = (48 - 50) / 100 = -0.02 mg per century.

The solution to the differential equation is:

m(t) = m(0) * e^(-kt)

We know that when t = 0, m(0) = 50 mg, so:

m(t) = 50 * e^(-0.02t)

Now, we want to find how long it takes for the sample to decay to 45 mg:

45 = 50 * e^(-0.02t)

Divide both sides by 50:

0.9 = e^(-0.02t)

Taking the natural logarithm of both sides:

ln(0.9) = -0.02t

Solve for t:

t = ln(0.9) / -0.02 ≈ 17.33 centuries

So, it will take approximately 17.33 centuries before there are 45 mg of the sample.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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