The half-lives of two radioactive materials A | vedclass

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(A) The mass of a radioactive substance remaining after time $t$ is given by $M = M_0 (1/2)^{t/T_{1/2}}$,where $M_0$ is the initial mass and $T_{1/2}$

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$2T$

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(A) The mass of a radioactive substance remaining after time $t$ is given by $M = M_0 (1/2)^{t/T_{1/2}}$,where $M_0$ is the initial mass and $T_{1/2}$ is the half-life.For material $A$: $M_A = M_{0A} (1/2)^{t/T}$.For material $B$: $M_B = M_{0B} (1/2)^{t/2T}$.Given the ratio of initial masses $M_{0A} / M_{0B} = 8/1$ and the ratio of remaining masses $M_A / M_B = 4/1$.Therefore,$\frac{M_A}{M_B} = \frac{M_{0A}}{M_{0B}} \cdot \frac{(1/2)^{t/T}}{(1/2)^{t/2T}} = 4/1$.Substituting the values: $8 \cdot (1/2)^{t/T - t/2T} = 4$.$8 \cdot (1/2)^{t/2T} = 4$.$(1/2)^{t/2T} = 4/8 = 1/2$.$(1/2)^{t/2T} = (1/2)^1$.Comparing the exponents,$t/2T = 1$,which gives $t = 2T$.

The half-lives of two radioactive materials $A$ and $B$ are $T$ and $2T$ respectively. If the ratio of the initial masses of the materials $A$ and $B$ is $8:1$,then the time after which the ratio of the masses of the materials $A$ and $B$ becomes $4:1$ is

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