Math Problem Statement

Plutonium is a radioactive atom, which decays at a rate of γ = (9.2) × 10^−10. Let p(t) denote the radioactivity at time t (t is in seconds) and p₀ the initial radioactivity. a) Find the inverse of this function and explain its meaning. b) How many years does it take for a piece of plutonium of initial radioactivity p₀ = 100 to decay to 5% of its initial radioactivity?

Solution

It seems you've uploaded an image with a question on radioactive decay involving a decay function. Let me extract and break down the problem for you.

The problem involves:

  1. A radioactive decay equation for Plutonium, p(t)=p0eγtp(t) = p_0 e^{-\gamma t}, where γ=9.2×1010\gamma = 9.2 \times 10^{-10}, tt is in seconds, p0p_0 is the initial radioactivity, and p(t)p(t) represents the radioactivity at time tt.
  2. You are asked to: a) Find the inverse of this function and explain its meaning. b) Calculate the time for Plutonium with an initial radioactivity of p0=100p_0 = 100 to decay to 5% of its initial radioactivity.

Let's solve each part step by step:

Part (a): Finding the inverse of p(t)=p0eγtp(t) = p_0 e^{-\gamma t}

We need to solve for tt in terms of p(t)p(t) and p0p_0.

Given: p(t)=p0eγtp(t) = p_0 e^{-\gamma t} To find the inverse, solve for tt:

  1. Divide both sides by p0p_0: p(t)p0=eγt\frac{p(t)}{p_0} = e^{-\gamma t}
  2. Take the natural logarithm (ln) of both sides to remove the exponential: ln(p(t)p0)=γt\ln \left( \frac{p(t)}{p_0} \right) = -\gamma t
  3. Solve for tt: t=ln(p(t)p0)γt = \frac{-\ln \left( \frac{p(t)}{p_0} \right)}{\gamma}

Thus, the inverse function is: t=ln(p(t)p0)γt = \frac{-\ln \left( \frac{p(t)}{p_0} \right)}{\gamma}

Meaning: The inverse function tt tells you the time it takes for the radioactivity to decay to a specified level p(t)p(t), given the initial radioactivity p0p_0 and the decay constant γ\gamma.

Part (b): Time for Plutonium to decay to 5% of its initial radioactivity

Here, you're given that p0=100p_0 = 100 and need to find the time tt when p(t)=0.05×p0p(t) = 0.05 \times p_0. This means: p(t)=5p(t) = 5 Using the inverse function from part (a): t=ln(p(t)p0)γt = \frac{-\ln \left( \frac{p(t)}{p_0} \right)}{\gamma} Substitute p(t)=5p(t) = 5, p0=100p_0 = 100, and γ=9.2×1010\gamma = 9.2 \times 10^{-10}:

t=ln(5100)9.2×1010t = \frac{-\ln \left( \frac{5}{100} \right)}{9.2 \times 10^{-10}} t=ln(0.05)9.2×1010t = \frac{-\ln (0.05)}{9.2 \times 10^{-10}} Now, let's calculate that value.

I'll perform this calculation.The time it takes for Plutonium to decay to 5% of its initial radioactivity is approximately 3.26×1093.26 \times 10^9 seconds, or about 103.4 years.

Would you like any further details or clarification on these calculations?

Here are five related questions you can explore:

  1. What is the half-life of Plutonium based on the given decay constant?
  2. How does the decay constant γ\gamma affect the rate of decay?
  3. What are the units of γ\gamma and how do they impact the equation?
  4. What is the radioactivity after 50 years for this Plutonium sample?
  5. Can this model be applied to other radioactive elements? How would their decay constants change?

Tip: Always check the units of each variable to ensure the results are consistent, especially in exponential decay problems.

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Math Problem Analysis

Mathematical Concepts

Exponential Decay
Inverse Functions

Formulas

p(t) = p₀e^{-γt}
Inverse: t = -ln(p(t) / p₀) / γ

Theorems

Exponential Decay Theorem
Logarithmic Properties

Suitable Grade Level

College level (Mathematics/Physics)