Question

{"given":"$$( t_{1/2} )_X = ( t_{\\text{mean}} )_Y, \\quad N_X(0) = N_Y(0) = N_0$$","key_observation":"Using $t_{1/2} = \\frac{\\ln 2}{\\lambda}$ and $t_{\\text{mean}} = \\frac{1}{\\lambda}$, setting them equal gives $\\lambda_Y = \\frac{\\lambda_X}{\\ln 2} = \\frac{\\lambda_X}{0.693}$, so $\\lambda_Y > \\lambda_X$. Since the initial number of atoms is the same for both and decay rate $= \\lambda N$, element Y has a higher initial decay rate.","option_analysis":[{"label":"(A)","text":"X and Y have the same decay rate initially.","verdict":"incorrect","explanation":"Since $\\lambda_Y > \\lambda_X$ and both start with the same $N_0$, their initial decay rates $\\lambda N_0$ are different — Y's is larger."},{"label":"(B)","text":"X and Y decay at the same rate always.","verdict":"incorrect","explanation":"They never have the same decay rate at any time since $\\lambda_Y > \\lambda_X$ throughout and they both decay exponentially with different constants."},{"label":"(C)","text":"Y will decay at a faster rate than X.","verdict":"correct","explanation":"From $(t_{1/2})_X = (t_{\\text{mean}})_Y$, we get $\\frac{\\ln 2}{\\lambda_X} = \\frac{1}{\\lambda_Y}$, so $\\lambda_Y = \\frac{\\lambda_X}{0.693} > \\lambda_X$. Hence, $\\lambda_Y N > \\lambda_X N$ at all times, meaning Y always decays faster."},{"label":"(D)","text":"X will decay at a faster rate than Y.","verdict":"incorrect","explanation":"Since $\\lambda_Y > \\lambda_X$, it is Y that decays faster, not X."}],"answer":"(C)","formula_steps":[]}