The half-life of a radioactive element A is the same as the mean-life of another radioactive element B. Initially both substances have the same number of atoms, then :

The half-life of element A, denoted as $$ T_{1/2,A} $$, is given by $$ T_{1/2,A} = \frac{\ln 2}{\lambda_A} $$, where $$ \lambda_A $$ is the decay constant for A. The mean-life of element B, denoted as $$ \tau_B $$, is given by $$ \tau_B = \frac{1}{\lambda_B} $$, where $$ \lambda_B $$ is the decay constant for B. The problem states that $$ T_{1/2,A} = \tau_B $$, so:

$$ \frac{\ln 2}{\lambda_A} = \frac{1}{\lambda_B} $$

Solving for $$ \lambda_B $$:

$$ \lambda_B = \frac{\lambda_A}{\ln 2} $$

Since $$ \ln 2 \approx 0.693 < 1 $$, it follows that:

$$ \lambda_B = \frac{\lambda_A}{0.693} \approx 1.443 \lambda_A $$

Thus, $$ \lambda_B > \lambda_A $$.

The decay rate (activity) at any time $$ t $$ is given by $$ R = \lambda N $$, where $$ N $$ is the number of atoms remaining. Initially, both A and B have the same number of atoms, say $$ N_0 $$. The initial decay rates are:

$$ R_A(0) = \lambda_A N_0 $$

$$ R_B(0) = \lambda_B N_0 = \frac{\lambda_A}{\ln 2} N_0 \approx 1.443 \lambda_A N_0 $$

Since $$ 1.443 > 1 $$, $$ R_B(0) > R_A(0) $$. Therefore, initially, B decays faster than A.

To compare the decay rates generally, consider the decay rates at time $$ t $$:

$$ N_A(t) = N_0 e^{-\lambda_A t} \quad \Rightarrow \quad R_A(t) = \lambda_A N_0 e^{-\lambda_A t} $$

$$ N_B(t) = N_0 e^{-\lambda_B t} \quad \Rightarrow \quad R_B(t) = \lambda_B N_0 e^{-\lambda_B t} = \frac{\lambda_A}{\ln 2} N_0 e^{-\frac{\lambda_A}{\ln 2} t} $$

The ratio of the decay rates is:

$$ \frac{R_B(t)}{R_A(t)} = \frac{\lambda_B}{\lambda_A} e^{-(\lambda_B - \lambda_A) t} = \frac{1}{\ln 2} e^{-\left( \frac{\lambda_A}{\ln 2} - \lambda_A \right) t} = \frac{1}{\ln 2} e^{-\lambda_A t \left( \frac{1}{\ln 2} - 1 \right)} $$

Simplify the exponent:

$$ \frac{1}{\ln 2} - 1 = \frac{1 - \ln 2}{\ln 2} $$

Since $$ \ln 2 \approx 0.693 $$, $$ 1 - \ln 2 \approx 0.307 > 0 $$, so the exponent is negative. Thus, the ratio $$ \frac{R_B(t)}{R_A(t)} $$ decreases over time from its initial value of approximately 1.443. However, at $$ t = 0 $$, $$ R_B > R_A $$, and since $$ \lambda_B > \lambda_A $$, B decays faster initially.

Evaluating the options:

• Option A: A and B decay at the same rate always. This is false because initially $$ R_B(0) > R_A(0) $$.
• Option B: A and B decay at the same rate initially. This is false because $$ R_B(0) \neq R_A(0) $$.
• Option C: A decays faster than B. This is false initially since $$ R_B(0) > R_A(0) $$, and while A may become faster later, the general statement does not hold.
• Option D: B decays faster than A. Given that $$ \lambda_B > \lambda_A $$ and initially $$ R_B(0) > R_A(0) $$, B decays faster than A initially and for some time thereafter. The decay constant $$ \lambda_B $$ being larger indicates B decays faster overall.

Hence, the correct answer is Option D.