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 determine the number of beta particles emitted by a sample of Co-60 in a given time period, we need to follow these steps:
Step 1: Determine the number of Co-60 nuclei in the sample.
To find the number of nuclei, we can use the formula:
Number of nuclei = (mass of sample / molar mass) * Avogadro's number
Given:
Mass of the sample (m) = 3.10 mg = 0.00310 g
Molar mass of Co-60 (M) = 59.93 g/mol (approximate value)
Avogadro's number (NA) = 6.02214076 × 10^23 mol^-1
Number of nuclei = (0.00310 g / 59.93 g/mol) * (6.02214076 × 10^23 mol^-1)
Step 2: Determine the decay constant (λ).
The decay constant is related to the half-life (T1/2) by the formula:
λ = ln(2) / T1/2
Given:
Half-life of Co-60 (T1/2) = 5.26 years
Decay constant (λ) = ln(2) / 5.26 years
Step 3: Determine the activity of the sample (A).
The activity (A) of a radioactive sample is given by the formula:
A = λ * N
where N is the number of nuclei.
Activity (A) = λ * Number of nuclei
Step 4: Determine the number of beta particles emitted in the given time period.
The number of beta particles depends on the electron emission probability (p). For Co-60, it undergoes beta decay and typically emits one beta particle.
Number of beta particles = p * A * t
Given:
Time period (t) = 190 seconds
Since Co-60 typically emits one beta particle, the electron emission probability (p) is 1.
Now let's apply the calculations:
Step 1:
Number of nuclei = (0.00310 g / 59.93 g/mol) * (6.02214076 × 10^23 mol^-1)
≈ 1.0038298 × 10^19 nuclei
Step 2:
Decay constant (λ) = ln(2) / 5.26 years
≈ 0.131683 yr^-1
Step 3:
Activity (A) = λ * Number of nuclei
≈ 0.131683 yr^-1 * 1.0038298 × 10^19 nuclei
Step 4:
Number of beta particles = p * A * t
= 1 * 0.131683 yr^-1 * 1.0038298 × 10^19 nuclei * 190 s
Once you substitute the known values into this equation, you can calculate the number of beta particles emitted during the given time period.
Additionally, to find the activity of the sample in becquerels (Bq), you can convert the activity from the previous step from per year to per second.
Activity (in Bq) = Activity (in yr^-1) / 3.7 × 10^10
Substitute the value of the calculated activity and follow the conversion to find the activity in becquerels.
To calculate the number of beta particles emitted in 190s, we need to first determine the number of cobalt-60 atoms present in the sample.
1. Calculate the number of moles in 3.10mg of cobalt-60:
- The molar mass of cobalt-60 (Co-60) is approximately 59.93 g/mol.
- Convert the mass to grams: 3.10mg = 0.00310g
- Convert grams to moles using the molar mass: 0.00310g / 59.93 g/mol = 5.172 x 10^-5 mol.
2. Use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms:
- Number of atoms = moles x Avogadro's number
- Number of atoms = (5.172 x 10^-5 mol) x (6.022 x 10^23 atoms/mol) = 3.114 x 10^19 atoms.
Now, let's calculate the number of beta particles emitted in 190s.
3. Determine the decay constant (λ):
- The decay constant (λ) can be calculated using the half-life (t1/2) formula: λ = ln(2) / t1/2
- λ = ln(2) / 5.26 yr
- Convert years to seconds: 5.26 yr x 365 days/yr x 24 hours/day x 60 min/hour x 60 s/min = 1.66 x 10^8 s
- λ = ln(2) / (1.66 x 10^8 s) = 4.18 x 10^-9 s^-1.
4. Calculate the number of beta particles emitted using the decay equation:
- Number of beta particles emitted (N) = initial number of atoms (N0) * (1 - e^(-λt))
- N = (3.114 x 10^19 atoms) * (1 - e^(-4.18 x 10^-9 s^-1 * 190 s))
- N ≈ 3.114 x 10^19 * (1 - e^(-0.0007942)) ≈ 3.114 x 10^19 * (1 - 0.9992059) ≈ 2.223 x 10^15 beta particles.
Finally, let's calculate the activity of the sample in Bq.
5. Activity (A) is defined as the number of radioactive decays per unit time (Bq):
- Activity (A) = λ * N
- A = (4.18 x 10^-9 s^-1) * (2.223 x 10^15 beta particles)
- A ≈ 9.27 x 10^6 Bq.
Therefore, approximately 2.223 x 10^15 beta particles are emitted in 190s by a 3.10mg sample of Co-60, and the activity of the sample is approximately 9.27 x 10^6 Bq.