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Math  /  Algebra

Question10. The half-life of the radioactive element plutonium-239 is 25,000 years. If 11 kilograms of plutonium-239 are initially present (between the size of a softball and shotput), how many years will it take for it to decay to less than 1 kilogram?

Studdy Solution

STEP 1

1. The half-life of plutonium-239 is 25,000 years.
2. Initially, there are 11 kilograms of plutonium-239.
3. We want to find the time it takes for the plutonium-239 to decay to less than 1 kilogram.

STEP 2

1. Understand the concept of half-life and exponential decay.
2. Set up the exponential decay formula.
3. Solve for the time it takes for the quantity to decay to less than 1 kilogram.

STEP 3

Understand the concept of half-life and exponential decay.
The half-life of a substance is the time it takes for half of the substance to decay. For plutonium-239, this is 25,000 years. The decay process follows an exponential pattern.

STEP 4

Set up the exponential decay formula.
The formula for exponential decay is:
N(t)=N0(12)tT N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}}
where: - N(t) N(t) is the quantity of substance remaining after time t t . - N0 N_0 is the initial quantity of the substance. - T T is the half-life of the substance.
For this problem: - N0=11 N_0 = 11 kilograms - T=25,000 T = 25,000 years - We want N(t)<1 N(t) < 1 kilogram

STEP 5

Solve for the time it takes for the quantity to decay to less than 1 kilogram.
We need to solve the inequality:
11(12)t25,000<1 11 \left(\frac{1}{2}\right)^{\frac{t}{25,000}} < 1
Divide both sides by 11:
(12)t25,000<111 \left(\frac{1}{2}\right)^{\frac{t}{25,000}} < \frac{1}{11}
Take the natural logarithm of both sides:
ln⁡((12)t25,000)<ln⁡(111) \ln\left(\left(\frac{1}{2}\right)^{\frac{t}{25,000}}\right) < \ln\left(\frac{1}{11}\right)
Using the property of logarithms, this becomes:
t25,000⋅ln⁡(12)<ln⁡(111) \frac{t}{25,000} \cdot \ln\left(\frac{1}{2}\right) < \ln\left(\frac{1}{11}\right)
Solve for t t :
t>ln⁡(111)ln⁡(12)×25,000 t > \frac{\ln\left(\frac{1}{11}\right)}{\ln\left(\frac{1}{2}\right)} \times 25,000
Calculate the right-hand side:
t>ln⁡(0.0909)ln⁡(0.5)×25,000 t > \frac{\ln(0.0909)}{\ln(0.5)} \times 25,000
t>−2.3979−0.6931×25,000 t > \frac{-2.3979}{-0.6931} \times 25,000
t>3.459×25,000 t > 3.459 \times 25,000
t>86,475 t > 86,475
Therefore, it will take more than 86,475 years for the plutonium-239 to decay to less than 1 kilogram.

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