Question

polonium 210 ($^{210}$po) has a half-life of approximately 138.6 days. if 4 g of $^{210}$po is present initially, then the amount of polonium 210 still present after t days is given by $a(t) = 4e^{-0.005t}$.

part 1 of 2

(a) find the amount of polonium 210 still present after 50 days. round to the nearest tenth of a gram.

the amount still present after 50 days is approximately \boxed{} g.

part 2 of 2

(b) find the amount of time necessary for the amount of $^{210}$po to decay to 0.4 g. round to the nearest tenth of a day.

the time necessary for the polonium 210 to decay to 0.4 g is approximately \boxed{} days.

Explanation:

Part 1 (a)

Step 1: Substitute \( t = 50 \) into the formula

The formula for the amount of polonium - 210 is \( A(t)=4e^{-0.005t} \). We need to find the amount after \( t = 50 \) days. So we substitute \( t = 50 \) into the formula: \( A(50)=4e^{-0.005\times50} \).
First, calculate the exponent: \( - 0.005\times50=-0.25 \). So the formula becomes \( A(50) = 4e^{-0.25} \).

Step 2: Calculate the value of \( e^{-0.25} \) and then multiply by 4

We know that \( e^{-0.25}=\frac{1}{e^{0.25}}\approx\frac{1}{1.284025417}\approx0.7788 \).
Then \( A(50)=4\times0.7788 = 3.1152 \). Rounding to the nearest tenth, we look at the hundredth place. The hundredth digit is 1, which is less than 5, so we round down. So \( A(50)\approx3.1 \).

Step 1: Set up the equation

We know that \( A(t) = 0.4 \) g and \( A(t)=4e^{-0.005t} \). So we set up the equation: \( 4e^{-0.005t}=0.4 \).

Step 2: Solve for \( t \)

First, divide both sides of the equation by 4: \( e^{-0.005t}=\frac{0.4}{4}=0.1 \).
Then, take the natural logarithm of both sides. Recall that if \( y = e^{x} \), then \( \ln(y)=x \). So taking \( \ln \) of both sides of \( e^{-0.005t}=0.1 \), we get \( \ln(e^{-0.005t})=\ln(0.1) \).
Since \( \ln(e^{x}) = x \), the left - hand side simplifies to \( - 0.005t \). So we have the equation \( -0.005t=\ln(0.1) \).
We know that \( \ln(0.1)\approx - 2.302585093 \). Then we solve for \( t \): \( t=\frac{\ln(0.1)}{-0.005}=\frac{- 2.302585093}{-0.005}=460.5170186 \).
Rounding to the nearest tenth, we get \( t\approx460.5 \).

Answer:

\( 3.1 \)

Part 2 (b)