Questions: The half-life of Palladium-100 is 4 days. After 20 days a sample of Palladium-100 has been reduced to a mass of 4 mg. What was the initial mass (in mg ) of the sample? What is the mass 8 weeks after the start?

Solution Steps

We are given the half-life of Palladium-100 as 4 days. We need to find the initial mass of the sample given that after 20 days, the mass is reduced to 4 mg. Additionally, we need to find the mass of the sample 8 weeks after the start.

The formula for exponential decay based on half-life is:

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

where:

• \( m(t) \) is the mass at time \( t \),
• \( m_0 \) is the initial mass,
• \( T_{1/2} \) is the half-life,
• \( t \) is the time elapsed.

Given:

• \( m(20) = 4 \) mg,
• \( T_{1/2} = 4 \) days,
• \( t = 20 \) days.

Substitute these values into the formula:

\[ 4 = m_0 \left(\frac{1}{2}\right)^{\frac{20}{4}} \]

Simplify the exponent:

\[ 4 = m_0 \left(\frac{1}{2}\right)^5 \]

Calculate \(\left(\frac{1}{2}\right)^5 = \frac{1}{32}\):

\[ 4 = m_0 \times \frac{1}{32} \]

Solve for \( m_0 \):

\[ m_0 = 4 \times 32 = 128 \text{ mg} \]

Convert 8 weeks to days:

\[ 8 \text{ weeks} = 8 \times 7 = 56 \text{ days} \]

Use the decay formula with \( t = 56 \) days:

\[ m(56) = 128 \left(\frac{1}{2}\right)^{\frac{56}{4}} \]

Simplify the exponent:

\[ m(56) = 128 \left(\frac{1}{2}\right)^{14} \]

Calculate \(\left(\frac{1}{2}\right)^{14} = \frac{1}{16384}\):

\[ m(56) = 128 \times \frac{1}{16384} \]

Calculate the mass:

\[ m(56) = \frac{128}{16384} = 0.0078125 \text{ mg} \]

Final Answer