Questions: Strontium-90 has a half-life of 28 days. Suppose a sample has an initial mass of 225 mg. a) Find an expression for the mass of the sample of Strontium-90 after t days. m(t)=225 * (1/2)^(t/28) b) Find the mass of the sample after 70 days. Round to three decimal places. c) How long does it take for the sample to contain only 3 mg of Strontium-90? Round to three decimal places.

Solution Steps

The mass of a radioactive substance that decays over time can be modeled using the exponential decay formula:

\[ m(t) = m_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T}} \]

where:

• \( m(t) \) is the mass at time \( t \),
• \( m_0 \) is the initial mass,
• \( T \) is the half-life of the substance.

For Strontium-90, \( m_0 = 225 \) mg and \( T = 28 \) days. Thus, the expression for the mass of the sample after \( t \) days is:

\[ m(t) = 225 \cdot \left(\frac{1}{2}\right)^{\frac{t}{28}} \]

To find the mass after 70 days, substitute \( t = 70 \) into the expression:

\[ m(70) = 225 \cdot \left(\frac{1}{2}\right)^{\frac{70}{28}} \]

Calculate the exponent:

\[ \frac{70}{28} = 2.5 \]

Now calculate the mass:

\[ m(70) = 225 \cdot \left(\frac{1}{2}\right)^{2.5} = 225 \cdot 0.1768 \approx 39.780 \]

To find the time \( t \) when the mass is 3 mg, set \( m(t) = 3 \) and solve for \( t \):

\[ 3 = 225 \cdot \left(\frac{1}{2}\right)^{\frac{t}{28}} \]

Divide both sides by 225:

\[ \frac{3}{225} = \left(\frac{1}{2}\right)^{\frac{t}{28}} \]

Simplify:

\[ 0.0133 = \left(\frac{1}{2}\right)^{\frac{t}{28}} \]

Take the logarithm of both sides:

\[ \log(0.0133) = \frac{t}{28} \cdot \log\left(\frac{1}{2}\right) \]

Solve for \( t \):

\[ t = 28 \cdot \frac{\log(0.0133)}{\log(0.5)} \]

Calculate:

\[ t \approx 28 \cdot \frac{-1.8761}{-0.3010} \approx 174.444 \]

Final Answer