Cobalt-60, which undergoes beta decay, has a half-life of 5.26 yr.
(a) How many beta particles are emitted in 47.0 s by a 2.39-mg sample of 60Co?
i got 2.44e19 but thast wrong
If you will post your work I'll try to find the error.
To find the number of beta particles emitted by a 2.39-mg sample of Cobalt-60 in 47.0 seconds, we can use the following steps:
First, we need to determine the number of Cobalt-60 atoms in the sample. To do this, we can use the concept of moles.
1. Calculate the molar mass of Cobalt-60 (60Co):
- Cobalt-60 has a molar mass of approximately 60 g/mol.
2. Convert the mass of the sample to moles:
- Divide the mass of the sample by the molar mass of Cobalt-60.
- In this case, divide 2.39 mg by 60 g/mol to obtain the number of moles.
3. Use Avogadro's number to convert moles to the number of atoms:
- Multiply the number of moles by Avogadro's number, which is approximately 6.022 × 10^23 atoms/mol.
- This will give you the number of Cobalt-60 atoms in the sample.
Now that we have the number of Cobalt-60 atoms, we can calculate the number of beta decay events that will occur in 47.0 seconds.
4. Determine the decay constant (λ):
- The decay constant (λ) is related to the half-life (t1/2) by the formula: λ = ln(2) / t1/2.
- In this case, the half-life of Cobalt-60 is given as 5.26 years. To convert years to seconds, multiply by 365 (days/year), 24 (hours/day), 60 (minutes/hour), and 60 (seconds/minute).
5. Calculate the number of beta decay events:
- Multiply the decay constant (λ) by the number of Cobalt-60 atoms in the sample and by the duration of time (47.0 seconds).
By following these steps and performing the calculations, you should be able to find the correct answer.