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(A) Let the initial mass of both radioactive elements $A$ and $B$ be $M_0$. Since their atomic weights are the same, the initial number of atoms $(N_0

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$\left(\frac{200}{7}\right) \ yrs$

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(A) Let the initial mass of both radioactive elements $A$ and $B$ be $M_0$. Since their atomic weights are the same, the initial number of atoms $(N_0)_A = (N_0)_B = N_0$.For element $A$, the amount remaining after time $t$ is $N_A = N_0 \left(\frac{1}{2}\right)^{t/10} = \frac{1}{e}$.For element $B$, the amount remaining after time $t$ is $N_B = N_0 \left(\frac{1}{2}\right)^{t/20} = 1$.Dividing the two equations:$\frac{N_A}{N_B} = \frac{1/e}{1} = \frac{N_0 (1/2)^{t/10}}{N_0 (1/2)^{t/20}}$$\frac{1}{e} = \left(\frac{1}{2}\right)^{t/10 - t/20} = \left(\frac{1}{2}\right)^{t/20}$.Taking natural logarithm on both sides:$\ln(e^{-1}) = \frac{t}{20} \ln(1/2)$$-1 = \frac{t}{20} (-\ln 2)$$1 = \frac{t}{20} (0.7)$$t = \frac{20}{0.7} = \frac{200}{7} \ yrs$.

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