91Ó°ÊÓ

To find the mass of the radioactive element, we need to determine the decay constant (λ). The decay constant is related to the half-life (T) using the following formula: \(λ = \frac{0.693}{T}\) Where \(T = 3.8\) hrs. Now, we can determine the decay constant: \(λ = \frac{0.693}{3.8 \mathrm{~hrs}}\) \(λ = 0.182 \mathrm{~hr}^{-1}\)

The activity (A) of the radioactive element is given as \(10^{16}\) Rutherford (1 Rutherford = 1 disintegration/s). We can determine the number of nuclei (N) using the radioactive activity and the decay constant. The activity formula is given by: \(A = λN\) We can rearrange the formula to find N: \(N = \frac{A}{λ}\) Now, let's find the number of nuclei (N): \(N = \frac{10^{16} \mathrm{~s}^{-1}}{0.182 \mathrm{~hr}^{-1}}\) \(N = \frac{10^{16}}{0.182 \times 3600 \mathrm{~s}^{-1}}\) \(N \approx 1.53 \times 10^{20}\) nuclei

Now, we have the number of nuclei (N). We can determine the mass (m) by multiplying N by the mass of one nucleus of \({ }^{222} R_{u}\). The atomic mass unit (u) is equal to approximately \(1.66 \times 10^{-27} \mathrm{~kg}\). The mass of one nucleus of \({ }^{222} R_{u}\) is \(222u\). Let's find the mass of \({ }^{222} R_{u}\): \(m = N \times (222u)\) \(m = (1.53 \times 10^{20}) \times (222 \times 1.66 \times 10^{-27} \mathrm{~kg})\) \(m \approx 6.74 \times 10^{-4} \mathrm{~kg}\) Convert the mass from kilograms to grams: \(m \approx 0.674 \mathrm{~g}\) The mass of \({ }^{222} R_{u}\) with activity equal to \(10^{16}\) Rutherford is approximately 0.674 g. Therefore, the correct choice is close to (B) 0.37 g, which is the closest option to our result. Note that there may be a discrepancy in the result due to approximations made in the calculations.