What is the activity of a one milligram sample of 210 Po after one year of decay, given its half-life of 138.3 days?

To find the activity of a one milligram sample of Polonium-210 (210 Po) after one year, we first need to determine how many half-lives have passed in that time period.

The half-life of 210 Po is 138.3 days. One year is approximately 365.25 days (considering leap years). To find the number of half-lives, we divide the total time by the half-life:

Number of half-lives = Total time / Half-life

Number of half-lives = 365.25 days / 138.3 days ≈ 2.64

This means that approximately 2.64 half-lives have passed in one year. Now we can calculate the remaining quantity of 210 Po using the formula:

N = N0 * (1/2)^(t/T)

Where:

• N0 = initial quantity of substance
• N = remaining quantity after time t
• T = half-life
• t = total time elapsed

Initially, we have 1 mg of 210 Po, which is:

N0 = 1 mg = 1 x 10^-3 g

Now we can calculate:

N = 1 x 10^-3 g * (1/2)^(2.64)

Calculating (1/2)^(2.64):

Approximately (1/2)^(2.64) ≈ 0.1875

So,

N ≈ 1 x 10^-3 g * 0.1875 ≈ 1.875 x 10^-4 g

Now, the activity (A) can be calculated with the formula:

A = λN

Where λ (decay constant) is given by:

λ = ln(2)/T

Using the half-life:

λ = ln(2)/138.3 days ≈ 0.00501 days^-1

We now calculate the activity:

A = λN = 0.00501 days^-1 * 1.875 x 10^-4 g

To convert grams to moles, we use the molar mass of 210 Po, which is about 210 g/mol. So:

N (in moles) = 1.875 x 10^-4 g / 210 g/mol ≈ 8.93 x 10^-7 mol

Using Avogadro’s number (6.022 x 10²³ atoms/mol) to find the number of atoms:

N (in atoms) = 8.93 x 10^-7 mol * 6.022 x 10²³ atoms/mol ≈ 5.37 x 10¹⁶ atoms

Finally, substituting this back to find the activity:

A = λ * N ≈ 0.00501 days^-1 * 5.37 x 10¹⁶ atoms ≈ 2.69 x 10¹⁴ disintegrations/day

Therefore, at the end of one year, the activity of the sample would be approximately 2.69 x 10¹⁴ disintegrations per day.