A sample of pure cobalt-60 has an activity of 8 µCi.
(t1/2 cobalt-60 = 5.26 a)
(a) How many atoms of cobalt-60 are present in the sample?
atoms
(b) What is the mass in grams of the sample?
g
The way I approached part a was to first convert the microCi into Bq (for a value of 296000 Bq). I then tried to find the decay constant, k, by converting 5.26a into seconds (for a value of 16.6E7 s) and using the equation ln2/((16.6E7) and ended up with 4.17E-9 (note: for the t I used more significant values then listed)....
After this I'm not so sure...but I tried using N=-dN/dT= (-2960000/(4.178E-9)= -7.083E12 atoms...which is of course wrong. Can someone redirect me in my way of thinking?
Thanks
I think your next step is Ro=kNo
The activity, Ro, = 296,000 dps
k = 4.178 x 10^-9 sec^-1, solve for No. Check my thinking.
Ah, the wonderful world of radioactive decay! Let me put on my clown nose and help you out with this question.
(a) To find the number of atoms of cobalt-60 present in the sample, we need to use Avogadro's constant and the molar mass of cobalt-60. However, since we know the activity of the sample, we can use the decay equation to solve for the number of atoms. The activity (A) is given by A = λN, where λ is the decay constant and N is the number of atoms.
In this case, the activity is 8 µCi, which is equivalent to 296,000 decays per second (since 1 Ci = 3.7 x 10^10 Bq). The decay constant (λ) can be found using λ = ln(2) / t1/2, where t1/2 is the half-life of cobalt-60.
Now, let's put everything together. Given that the activity (A) is 296,000 decays per second and the half-life (t1/2) of cobalt-60 is 5.26 years, we have:
296,000 = λN, and λ = ln(2) / 5.26.
Rearranging the equation, we can solve for N (the number of atoms):
N = 296,000 / λ.
Pop in the values and let the clown math do its magic!
(b) Now, let's move on to finding the mass in grams of the sample. We can use the molar mass of cobalt-60 and the number of atoms we found in part (a) to calculate the mass.
The molar mass of cobalt-60 is approximately 59.933 g/mol. To find the mass in grams, we can use the equation:
mass = N × (molar mass / Avogadro's constant).
Substitute the values and watch the clown science unfold!
I hope this helps to redirect your thinking in a fun and informative way. Remember, physics can be a circus sometimes, but with a little humor, it all falls into place!
To calculate the number of atoms of cobalt-60 present in the sample, you need to use the formula for radioactive decay:
N(t) = N(0) * e^(-λt)
where N(t) is the number of atoms at time t, N(0) is the initial number of atoms, λ is the decay constant, and t is the time.
(a) To find the number of atoms of cobalt-60 present in the sample, you can rearrange the equation as:
N(0) = N(t) / e^(-λt)
Given that the activity of the sample is 8 µCi (which is equivalent to 296,000 Bq), you can find the decay constant (λ) using the half-life of cobalt-60:
λ = ln(2) / (t1/2)
λ = ln(2) / (5.26*10^9 a * 365.25 days/a * 24 hours/day * 3600 seconds/hour)
Substituting the values and calculating λ, you'll get approximately 2.09 * 10^(-12) s^(-1).
Now, we can find the initial number of atoms (N(0)) by rearranging the equation as mentioned earlier:
N(0) = (Activity / λ)
N(0) = (296,000 Bq) / (2.09 * 10^(-12) s^(-1))
Calculating this will give you the number of atoms of cobalt-60 in the sample.
(b) To find the mass of the sample, you'll need to use the atomic mass of cobalt-60, which is approximately 59.933 atomic mass units (amu).
The mass of the sample can be calculated using the formula:
Mass = (N(0) * atomic mass) / Avogadro's number
Substituting the values, you'll get the mass of the sample in grams.
To solve part (a), let's start by finding the number of atoms of cobalt-60 in the sample.
The activity of cobalt-60 is usually expressed in becquerels (Bq) rather than microcuries (µCi). So, let's convert the activity from µCi to Bq.
1 µCi = 37,000 Bq
Therefore, the activity of the sample is 8 µCi * 37,000 Bq/µCi = 296,000 Bq.
Now, we can use the equation for radioactive decay to find the number of atoms:
A = λN
Where A is the activity, λ is the decay constant, and N is the number of atoms.
The decay constant, λ, can be calculated using the half-life (t1/2) of cobalt-60:
λ = ln(2) / t1/2
λ = ln(2) / 5.26 years = 0.1319 years⁻¹
To convert years⁻¹ to seconds⁻¹, we need to multiply by the conversion factor 3.156 x 10^7 seconds/year.
λ = 0.1319 years⁻¹ * (3.156 x 10^7 seconds/year) ≈ 4.16 x 10⁻⁹ seconds⁻¹
Now, we can rearrange the equation A = λN to solve for N:
N = A / λ
N = 296,000 Bq / (4.16 x 10⁻⁹ seconds⁻¹)
N ≈ 7.12 x 10¹³ atoms of cobalt-60
So, there are approximately 7.12 x 10¹³ atoms of cobalt-60 in the sample.
For part (b), to find the mass of the sample, we need to know the atomic mass of cobalt-60.
The atomic mass of cobalt-60 is 59.933822 g/mol.
Since cobalt-60 has one cobalt atom per molecule, the molar mass is the same as the atomic mass.
Now, we can calculate the mass of the sample:
Mass = N * Molar mass
Mass = (7.12 x 10¹³ atoms) * (59.933822 g/mol)
Mass ≈ 4.27 x 10¹⁵ g
So, the mass of the sample is approximately 4.27 x 10¹⁵ grams.