Calculate the mass of `.^(140)La` in a sample whose activity is `3.7xx10^(10)Bq` (`1` Becquerel, `Bq= 1` disintegration per second) given that is `t_(1//2)` is 40 hour.

Explanation

To calculate the mass of \(^{140}\text{La}\) in a sample with an activity of \(3.7 \times 10^{10} \text{ Bq}\) and a half-life of 40 hours, we can 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 in seconds. First, we need to convert the half-life from hours to seconds.

\[ t_{1/2} = 40 \text{ hours} \times 3600 \text{ seconds/hour} = 144000 \text{ seconds} \]

Now, substituting this value into the decay constant formula:
\[ \lambda = \frac{0.693}{144000} \approx 4.81 \times 10^{-6} \text{ s}^{-1} \]

### Step 2: Relate Activity to the Number of Nuclei
The activity \(A\) of a radioactive sample is given by the equation:
\[ A = \lambda N \]
where \(N\) is the number of radioactive nuclei. Rearranging this gives:
\[ N = \frac{A}{\lambda} \]

Substituting the values:
\[ N = \frac{3.7 \times 10^{10} \text{ Bq}}{4.81 \times 10^{-6} \text{ s}^{-1}} \approx 7.69 \times 10^{15} \text{ nuclei} \]

### Step 3: Calculate the Mass of \(^{140}\text{La}\)
The mass of the radioactive substance can be calculated using the formula:
\[ \text{mass} = \frac{N \times A}{N_A} \]
where \(N_A\) is Avogadro's number (\(6.022 \times 10^{23} \text{ mol}^{-1}\)) and \(A\) is the mass number of the isotope.

First, we need to convert the number of nuclei to moles:
\[ \text{moles} = \frac{N}{N_A} = \frac{7.69 \times 10^{15}}{6.022 \times 10^{23}} \approx 1.28 \times 10^{-8} \text{ moles} \]

Now, we can calculate the mass:
\[ \text{mass} = \text{moles} \times \text{molar mass} = 1.28 \times 10^{-8} \text{ moles} \times 140 \text{ g/mol} \approx 1.79 \times 10^{-6} \text{ g} \]

### Final Result
The mass of \(^{140}\text{La}\) in the sample is approximately \(1.79 \times 10^{-6} \text{ g}\).

---