Calculate the weight of `""^(14)C (t_((1)/(2)) =5720yr)` atoms which will give `3.70 xx 10^(7)` disintegrations per second (dps)

Answer

To solve the problem of calculating the weight of \(^{14}C\) atoms that will give \(3.70 \times 10^7\) disintegrations per second (dps), we will follow these steps:

### Step 1: Calculate the decay constant (\(\lambda\))
The decay constant \(\lambda\) can be calculated using the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} \]
where \(t_{1/2}\) is the half-life of the isotope. Given that the half-life of \(^{14}C\) is \(5720\) years, we first need to convert this into seconds.

\[ t_{1/2} = 5720 \text{ years} \times 365 \text{ days/year} \times 24 \text{ hours/day} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} \]
Calculating this gives:
\[ t_{1/2} = 5720 \times 365 \times 24 \times 60 \times 60 = 1.80 \times 10^{11} \text{ seconds} \]

Now substituting this value into the decay constant formula:
\[ \lambda = \frac{0.693}{1.80 \times 10^{11}} \approx 3.84 \times 10^{-12} \text{ s}^{-1} \]

### Step 2: Calculate the number of atoms required for the given disintegration rate
The relationship between the decay constant, the number of atoms \(N\), and the disintegration rate \(R\) is given by:
\[ R = \lambda N \]
Rearranging this gives:
\[ N = \frac{R}{\lambda} \]
Substituting the values:
\[ N = \frac{3.70 \times 10^7}{3.84 \times 10^{-12}} \approx 9.64 \times 10^{18} \text{ atoms} \]

### Step 3: Calculate the mass of \(^{14}C\) corresponding to the number of atoms
The mass of a single atom of \(^{14}C\) can be calculated using its molar mass and Avogadro's number:
\[ \text{Molar mass of } ^{14}C = 14 \text{ g/mol} \]
\[ \text{Mass of one atom} = \frac{14 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}} \approx 2.33 \times 10^{-26} \text{ g/atom} \]

Now, the total mass \(m\) of \(N\) atoms is given by:
\[ m = N \times \text{mass of one atom} \]
Substituting the values:
\[ m = 9.64 \times 10^{18} \times 2.33 \times 10^{-26} \approx 2.25 \times 10^{-7} \text{ g} \]

### Step 4: Convert the mass to milligrams
To convert grams to milligrams:
\[ m = 2.25 \times 10^{-7} \text{ g} \times 1000 \text{ mg/g} = 2.25 \times 10^{-4} \text{ mg} \]

### Final Answer
The weight of \(^{14}C\) atoms that will give \(3.70 \times 10^7\) disintegrations per second is approximately:
\[ \boxed{2.25 \times 10^{-4} \text{ mg}} \]