Let’s break down the solution to the given problem:
### a) Number of Radium Atoms in One Gram
1. Understanding the Atomic Mass and Avogadro's Number:
- The atomic mass of radium is 226 atomic mass units (amu).
- Avogadro's number, which is the number of atoms in one mole of any substance, is [tex]\(6.022 \times 10^{23}\)[/tex] atoms/mol.
2. Calculating Moles in One Gram of Radium:
- One mole of radium (which weighs 226 grams) contains [tex]\(6.022 \times 10^{23}\)[/tex] atoms.
- To find the number of moles in one gram of radium:
[tex]\[
\text{Moles in one gram} = \frac{1 \text{ gram}}{226 \text{ grams/mol}} = \frac{1}{226} \text{ mol}
\][/tex]
3. Calculating Number of Atoms in One Gram of Radium:
- Using the number of moles in one gram, the number of atoms can be found by multiplying by Avogadro's number:
[tex]\[
\text{Number of atoms} = \left(\frac{1}{226} \text{ mol}\right) \times 6.022 \times 10^{23} \text{ atoms/mol} = 2.6646 \times 10^{21} \text{ atoms}
\][/tex]
### b) Calculate the Disintegration Rate (Activity) Per Second
1. Convert Half-life to Seconds:
- The half-life of radium-226 is given as 1600 years.
- Converting years to seconds (considering a year has approximately 365.25 days):
[tex]\[
\text{Half-life in seconds} = 1600 \text{ years} \times 365.25 \text{ days/year} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} = 5.05 \times 10^{10} \text{ seconds}
\][/tex]
2. Calculate the Decay Constant (λ):
- The decay constant, λ, is related to the half-life [tex]\( T_{1/2} \)[/tex] by the formula:
[tex]\[
\lambda = \frac{\ln(2)}{T_{1/2}}
\][/tex]
- Plugging in the half-life in seconds:
[tex]\[
\lambda = \frac{\ln(2)}{5.05 \times 10^{10} \text{ seconds}} = 1.3728 \times 10^{-11} \text{ s}^{-1}
\][/tex]
3. Calculate the Activity (A) in Disintegrations per Second:
- Activity is given by the formula:
[tex]\[
A = \lambda \times N
\][/tex]
- Where [tex]\( N \)[/tex] is the number of atoms in one gram of radium we calculated previously:
[tex]\[
N = 2.6646 \times 10^{21} \text{ atoms}
\][/tex]
- So, the activity (A) is:
[tex]\[
A = 1.3728 \times 10^{-11} \text{ s}^{-1} \times 2.6646 \times 10^{21} \text{ atoms}
\][/tex]
[tex]\[
A = 3.6579 \times 10^{10} \text{ disintegrations per second}
\][/tex]
In summary, for the given problem:
a) The number of radium atoms in one gram is approximately [tex]\(2.6646 \times 10^{21}\)[/tex].
b) The disintegration rate, or activity, is approximately [tex]\(3.6579 \times 10^{10}\)[/tex] disintegrations per second.
