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### Part a) Amount of atoms present in 20 mg of Barium (2 marks)
1. Given values:
- Avogadro's constant ([tex]\(N_A\)[/tex]) = [tex]\(6 \times 10^{23}\)[/tex] atoms/mol
- Molar mass of Barium-142 ([tex]\(M_{\text{Ba}}\)[/tex]) = 142 g/mol
- Initial mass of Barium ([tex]\(m_{\text{initial}}\)[/tex]) = 20 mg
2. Convert the initial mass from mg to grams:
[tex]\[ m_{\text{initial}} = \frac{20 \text{ mg}}{1000} = 0.02 \text{ g} \][/tex]
3. Calculate the number of moles of Barium:
[tex]\[ \text{Moles of Barium} = \frac{m_{\text{initial}}}{M_{\text{Ba}}} = \frac{0.02 \text{ g}}{142 \text{ g/mol}} = 1.408450704225352 \times 10^{-4} \text{ mol} \][/tex]
4. Calculate the number of atoms:
[tex]\[ \text{Atoms of Barium} = \text{Moles of Barium} \times N_A = 1.408450704225352 \times 10^{-4} \text{ mol} \times 6 \times 10^{23} \text{ atoms/mol} \][/tex]
[tex]\[ \text{Atoms of Barium} = 8.450704225352114 \times 10^{19} \text{ atoms} \][/tex]
### Part b) Activity of the Barium in Bq (3 marks)
1. Given values:
- Half-life ([tex]\(t_{1/2}\)[/tex]) = 60 minutes
2. Calculate the decay constant ([tex]\(\lambda\)[/tex]):
[tex]\[ \lambda = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{60} \text{ per minute} \approx 0.01155 \text{ per minute} \][/tex]
3. Calculate the initial activity ([tex]\(A_0\)[/tex]):
[tex]\[ A_0 = \lambda \times \text{Number of atoms} \][/tex]
[tex]\[ A_0 = 0.01155 \text{ per minute} \times 8.450704225352114 \times 10^{19} \text{ atoms} \][/tex]
[tex]\[ A_0 = 9.762636345914724 \times 10^{17} \text{ Bq} \][/tex]
### Part c) Barium's activity after the 20th half-life in Bq (3 marks)
1. Given values:
- Number of half-lives = 20
2. Calculate the remaining activity after 20 half-lives ([tex]\(A\)[/tex]):
[tex]\[ A = A_0 \times \left(\frac{1}{2}\right)^{20} \][/tex]
[tex]\[ A = 9.762636345914724 \times 10^{17} \text{ Bq} \times (\frac{1}{2})^{20} \][/tex]
[tex]\[ A \approx 9.31037554351.3035 \text{ Bq} \][/tex]
### Summary
- Step a) The number of atoms present in 20 mg of Barium is approximately [tex]\(8.450704225352114 \times 10^{19}\)[/tex] atoms.
- Step b) The initial activity of the Barium is approximately [tex]\(9.762636345914724 \times 10^{17}\)[/tex] Bq.
- Step c) The activity of Barium after 20 half-lives is approximately [tex]\(931037554351.3035\)[/tex] Bq.
### Part a) Amount of atoms present in 20 mg of Barium (2 marks)
1. Given values:
- Avogadro's constant ([tex]\(N_A\)[/tex]) = [tex]\(6 \times 10^{23}\)[/tex] atoms/mol
- Molar mass of Barium-142 ([tex]\(M_{\text{Ba}}\)[/tex]) = 142 g/mol
- Initial mass of Barium ([tex]\(m_{\text{initial}}\)[/tex]) = 20 mg
2. Convert the initial mass from mg to grams:
[tex]\[ m_{\text{initial}} = \frac{20 \text{ mg}}{1000} = 0.02 \text{ g} \][/tex]
3. Calculate the number of moles of Barium:
[tex]\[ \text{Moles of Barium} = \frac{m_{\text{initial}}}{M_{\text{Ba}}} = \frac{0.02 \text{ g}}{142 \text{ g/mol}} = 1.408450704225352 \times 10^{-4} \text{ mol} \][/tex]
4. Calculate the number of atoms:
[tex]\[ \text{Atoms of Barium} = \text{Moles of Barium} \times N_A = 1.408450704225352 \times 10^{-4} \text{ mol} \times 6 \times 10^{23} \text{ atoms/mol} \][/tex]
[tex]\[ \text{Atoms of Barium} = 8.450704225352114 \times 10^{19} \text{ atoms} \][/tex]
### Part b) Activity of the Barium in Bq (3 marks)
1. Given values:
- Half-life ([tex]\(t_{1/2}\)[/tex]) = 60 minutes
2. Calculate the decay constant ([tex]\(\lambda\)[/tex]):
[tex]\[ \lambda = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{60} \text{ per minute} \approx 0.01155 \text{ per minute} \][/tex]
3. Calculate the initial activity ([tex]\(A_0\)[/tex]):
[tex]\[ A_0 = \lambda \times \text{Number of atoms} \][/tex]
[tex]\[ A_0 = 0.01155 \text{ per minute} \times 8.450704225352114 \times 10^{19} \text{ atoms} \][/tex]
[tex]\[ A_0 = 9.762636345914724 \times 10^{17} \text{ Bq} \][/tex]
### Part c) Barium's activity after the 20th half-life in Bq (3 marks)
1. Given values:
- Number of half-lives = 20
2. Calculate the remaining activity after 20 half-lives ([tex]\(A\)[/tex]):
[tex]\[ A = A_0 \times \left(\frac{1}{2}\right)^{20} \][/tex]
[tex]\[ A = 9.762636345914724 \times 10^{17} \text{ Bq} \times (\frac{1}{2})^{20} \][/tex]
[tex]\[ A \approx 9.31037554351.3035 \text{ Bq} \][/tex]
### Summary
- Step a) The number of atoms present in 20 mg of Barium is approximately [tex]\(8.450704225352114 \times 10^{19}\)[/tex] atoms.
- Step b) The initial activity of the Barium is approximately [tex]\(9.762636345914724 \times 10^{17}\)[/tex] Bq.
- Step c) The activity of Barium after 20 half-lives is approximately [tex]\(931037554351.3035\)[/tex] Bq.
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