1 g of Cs decays with a half life of 30 years . If the initial activity of Cs is 1.0 mc(millicurie) , find the number of atoms left after 5 years and the activity after 5 years. Avogadro number is 6.02*10^23 per mole.

To solve the problem regarding the decay of cesium (Cs) with a half-life of 30 years, we need to break it down into a few logical steps. We will first determine the number of atoms present initially, then calculate how many atoms remain after 5 years, and finally, we will find the activity after that time period.

Step 1: Initial Number of Atoms

We start with 1 gram of Cs. To find the number of atoms, we first need to convert grams to moles using the molar mass of cesium. The molar mass of Cs is approximately 132.91 g/mol.

• Calculate the number of moles of Cs:

Number of moles = mass (g) / molar mass (g/mol) = 1 g / 132.91 g/mol ≈ 0.00752 moles.

• Now, we convert moles to atoms using Avogadro's number:

Number of atoms = moles × Avogadro's number = 0.00752 moles × 6.02 × 10²³ atoms/mole ≈ 4.52 × 10²¹ atoms.

Step 2: Calculate Remaining Atoms After 5 Years

Next, we need to determine how many of these atoms remain after 5 years. The decay of radioactive substances follows an exponential decay model, which can be expressed with the formula:

N(t) = N₀ × (1/2)^(t/T₁/₂)

Where:

• N(t) = number of atoms remaining after time t
• N₀ = initial number of atoms
• T₁/₂ = half-life of the substance
• t = time elapsed

In this case, the half-life (T₁/₂) is 30 years, and we want to find the number of atoms after 5 years (t = 5 years).

Plugging in the values:

N(5) = 4.52 × 10²¹ × (1/2)^(5/30)

Calculating the exponent:

(1/2)^(5/30) = (1/2)^(1/6) ≈ 0.8909

Now, substituting back into the equation:

N(5) ≈ 4.52 × 10²¹ × 0.8909 ≈ 4.02 × 10²¹ atoms.

Step 3: Determine Activity After 5 Years

Activity (A) is defined as the number of decays per unit time, and it can be calculated using the formula:

A = λN

Where λ (decay constant) can be calculated from the half-life using the formula:

λ = ln(2) / T₁/₂

Calculating λ:

λ = 0.693 / 30 years ≈ 0.0231 year⁻¹.

Now, we can find the activity after 5 years using the number of remaining atoms:

A(5) = λ × N(5) = 0.0231 year⁻¹ × 4.02 × 10²¹ atoms ≈ 9.30 × 10²⁰ decays/year.

To convert this into millicuries (mCi), we use the conversion factor where 1 Ci = 3.7 × 10¹⁰ decays/second. Therefore:

A(5) in Ci = 9.30 × 10²⁰ decays/year ÷ (3.7 × 10¹⁰ decays/second × 31,536,000 seconds/year) ≈ 0.00029 Ci.

Converting to millicuries:

A(5) in mCi = 0.00029 Ci × 1000 mCi/Ci ≈ 0.29 mCi.

Summary of Results

After 5 years, the number of cesium atoms remaining is approximately 4.02 × 10²¹ atoms, and the activity is about 0.29 mCi. This illustrates how radioactive decay works over time and how to apply the concepts of half-life and activity in practical calculations.