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?
k = 0.693/t1/2
k will be in yr^-1.
Convert 310 mg Co to g, then to moles. moles = grams/atomic mass.
moles x 6.02E23 = # atoms.
Rate = # atoms x k(from above) = decays per year.
Convert that to 190 sec.
Convert that to decays per sec (dps) and that will be # Bq.
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To find the number of beta particles emitted in 190 seconds by a 3.10 mg sample of Co-60, we first need to calculate the number of Co-60 atoms in the sample. Then, we can use the decay constant and the decay equation to determine the number of beta particles emitted.
1. Calculate the number of Co-60 atoms in the sample:
- First, convert the mass of the sample from milligrams (mg) to grams (g):
3.10 mg = 3.10 × 10^-3 g
- Next, use the molar mass of Co-60 to convert grams to moles:
Molar mass of Co-60 = 59.93 g/mol
Number of moles of Co-60 = (3.10 × 10^-3 g) / (59.93 g/mol)
- Finally, use Avogadro's number (6.022 × 10^23 atoms/mol) to convert moles to atoms:
Number of Co-60 atoms = Number of moles of Co-60 × Avogadro's number
2. Calculate the number of beta particles emitted:
- Use the decay constant equation: λ = ln(2) / t1/2
Decay constant (λ) = ln(2) / 5.26 yr
Decay constant (λ) = [ln(2) / 5.26 yr] × [1 yr / 365 days] × [1 day / 24 hours] × [1 hour / 3600 seconds] × [1 second]
- Finally, use the decay equation: N = N0 * e^(-λt)
Number of beta particles emitted = N0 - N
Number of beta particles emitted = N0 * (1 - e^(-λt))
3. Calculate the activity of the sample in Bq:
- Activity (A) = λ * N0
Now, we can calculate each step:
Step 1:
Number of moles of Co-60 = (3.10 × 10^-3 g) / (59.93 g/mol)
Number of Co-60 atoms = Number of moles of Co-60 × Avogadro's number
Step 2:
Decay constant (λ) = [ln(2) / 5.26 yr] × [1 yr / 365 days] × [1 day / 24 hours] × [1 hour / 3600 seconds] × [1 second]
Number of beta particles emitted = Number of Co-60 atoms * (1 - e^(-λt))
Step 3:
Activity (A) = λ * Number of Co-60 atoms
Now, let's calculate the values step by step.
To determine the number of beta particles emitted in 190 seconds by a sample of Co-60, we can use the radioactive decay formula:
N(t) = N(0) * (1/2)^(t / T)
Where:
N(t) is the number of radioactive atoms remaining after time t,
N(0) is the initial number of radioactive atoms,
t is the elapsed time,
T is the half-life of the substance.
First, let's find the initial number of radioactive atoms (N(0)) in the 3.10mg sample of Co-60.
1. Determine the number of moles of Co-60:
Number of moles (n) = mass (m) / molar mass (M)
Given mass = 3.10mg
Molar mass of Co-60 is approximately 59.933 g/mol.
n = 3.10mg / 59.933 g/mol
n = 0.0517 mol
2. Find the Avogadro's number (N) since it represents the number of atoms in one mole:
N = 6.022 x 10^23 particles/mol
Number of radioactive atoms (N(0)) = n * N
N(0) = 0.0517 mol * 6.022 x 10^23 particles/mol
Now, substitute the values into the radioactive decay equation to find N(t):
N(t) = N(0) * (1/2)^(t / T)
N(t) = (0.0517 mol * 6.022 x 10^23 particles/mol) * (1/2)^(190s / 5.26yr)
Convert years to seconds:
1 year = 365 days * 24 hours * 60 minutes * 60 seconds
1 year = 31,536,000 seconds
N(t) = (0.0517 mol * 6.022 x 10^23 particles/mol) * (1/2)^(190s / 5.26 * 31,536,000s)
Simplifying this equation will give you the number of beta particles emitted in 190 seconds by the Co-60 sample.
To calculate the activity of the sample in Bq (Becquerel), we need to find the decay constant (λ) first, which can be derived from the half-life (T):
λ = ln(2) / T
Substituting the known value of T (5.26 years) into the equation will give you the decay constant (λ).
Once you have the decay constant (λ), you can calculate the activity (A) using the following formula:
A = λ * N(0)
Substitute the values into the formula to find the activity of the Co-60 sample in Bq.