The half life of cesium-137 is 30 years. Suppose we have a 10-g sample.

a)Find a function that models the mass remaining after t years.
b)How much of the sample will remain after 80 years?
c)After how long will only 2 g of the sample remain?

Thanks

(a) M(t) = mass of Cesium-137 at time t, in years = Mo* 2^(-t/30)
Mo is the initial mass of 10 g.
2^(-t/30) means 2 to the -t/30 power

(b) Plug in t = 80 and see that you get for M

(c) Set M = 2 and see what you get for t. Hint: take logs of both sides of:
2 = 10 * 2^(-t/30)
(0.2) = 2^(-t/30)

To find a function that models the mass remaining after time t, we can use the exponential decay formula. The formula for exponential decay is:

M(t) = Mo * e^(-λt)

Where:
M(t) is the mass remaining after time t
Mo is the initial mass
λ is the decay constant
t is the time

In this case, we have the half-life of cesium-137, which is 30 years. The decay constant (λ) can be calculated using the formula:

λ = ln(2) / half-life

Substituting the values into the formula:

λ = ln(2) / 30

Now we can write the function that models the mass remaining after t years:

M(t) = Mo * e^(-ln(2) / 30 * t)

For part (b), we need to find the amount of the sample remaining after 80 years. We can plug t = 80 into the function to get the result:

M(80) = 10 * e^(-ln(2) / 30 * 80)

For part (c), we want to know how long it will take for only 2 g of the sample to remain. We can set M(t) = 2 and solve for t:

2 = 10 * e^(-ln(2) / 30 * t)

To solve this equation, we can take the natural logarithm (ln) of both sides:

ln(2) = -ln(2) / 30 * t * ln(e)

Simplifying:

ln(2) = -t / 30 * ln(2)

Now we can solve for t:

t = ln(2) / (30 * ln(2))

Note: ln(2) is approximately 0.693147

Substituting the value of ln(2) into the equation:

t = 0.693147 / (30 * 0.693147)

Simplifying further:

t = 1 / 30

Therefore, after approximately 1/30 of a year (or about 0.0333 years), only 2 g of the sample will remain.

(a) The function that models the mass remaining after t years is M(t) = 10 * 2^(-t/30). Here, M(t) represents the mass of Cesium-137 at time t in years, and 10 is the initial mass of the sample.

(b) To find out how much of the sample will remain after 80 years, we can plug in t = 80 into the function and calculate M(80).

M(80) = 10 * 2^(-80/30)
M(80) ≈ 10 * 2^(-8/3)
M(80) ≈ 10 * 2^(-2.67)
M(80) ≈ 10 * 0.138
M(80) ≈ 1.38 g

So, approximately 1.38 grams of the sample will remain after 80 years.

(c) To find out after how long only 2 grams of the sample will remain, we need to solve the equation 2 = 10 * 2^(-t/30).

Taking the logarithm of both sides, we get:
log(2) = log(10 * 2^(-t/30))
log(2) = log(10) + log(2^(-t/30))
log(2) = log(10) - (t/30)log(2)

Now, we can solve for t:
(t/30)log(2) = log(10) - log(2)
(t/30)log(2) = log(10/2)
(t/30)log(2) = log(5)
t/30 = log(5) / log(2)
t = (30 * log(5)) / log(2)

Using a calculator, we can approximate the value of t as:
t ≈ 101.43 years

So, after approximately 101.43 years, only 2 grams of the sample will remain.