Cobalt-60, which undergoes beta decay, has a half-life of 5.26 yr. How many beta particles are emitted in 190s by a 3.10mg sample of Co-60? What is the activity of the sample in Bq?
To find the number of beta particles emitted in 190s by a 3.10mg sample of Co-60, we need to use the radioactive decay formula:
N(t) = N(0) * (1/2)^(t / half-life)
Where:
- N(t) is the number of atoms remaining after time t
- N(0) is the initial number of atoms
- t is the time elapsed
- half-life is the time it takes for half the atoms to decay
Step 1: Find the initial number of atoms (N(0))
To find the initial number of atoms in the sample, we need to use the Avogadro's number (6.022 × 10^23) and the molar mass of Co-60.
1 mole of an element contains Avogadro's number of atoms. The molar mass of Co-60 is 60g/mol, which means that 1 mole of Co-60 weighs 60g and contains 6.022 × 10^23 atoms.
Convert the mass of the sample to moles:
Moles = Mass / Molar mass
Moles = 3.10mg / 60g/mol
Moles = 0.0517 mol
Calculate the number of atoms in the sample:
Number of atoms = Moles * Avogadro's number
Number of atoms = 0.0517 mol * 6.022 × 10^23 atoms/mol
Number of atoms = 3.113 × 10^22 atoms (approximately)
Step 2: Find the number of atoms remaining after 190s (N(t))
Using the half-life, we can calculate the number of atoms remaining after 190s:
N(t) = N(0) * (1/2)^(t / half-life)
N(t) = 3.113 × 10^22 * (1/2)^(190s / 5.26yr)
Before continuing, we need to convert 190s to years:
190s * (1 min / 60s) * (1 hr / 60 min) * (1 day / 24 hr) * (1 yr / 365 days) ≈ 0.00602 yr
Now, substitute the values:
N(t) = 3.113 × 10^22 * (1/2)^(0.00602 yr / 5.26yr)
N(t) ≈ 3.067 × 10^22 atoms
Step 3: Find the number of beta particles emitted
The number of beta particles emitted is equal to the difference between the initial number of atoms and the number of atoms remaining:
Number of beta particles emitted = N(0) - N(t)
Number of beta particles emitted = 3.113 × 10^22 - 3.067 × 10^22
Number of beta particles emitted ≈ 4.6 × 10^20 beta particles
To calculate the activity of the sample in Bq, we use the definition that one becquerel (Bq) is one decay event per second.
Activity = Number of beta particles emitted / Time (in seconds)
Activity = 4.6 × 10^20 / 190
Activity ≈ 2.42 × 10^18 Bq
Therefore, approximately 4.6 × 10^20 beta particles are emitted in 190s by a 3.10mg sample of Co-60, and the activity of the sample is approximately 2.42 × 10^18 Bq.