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(B) The half-life $T_{1/2}$ is the time taken for a substance to reduce to half its initial mass.Given that $20 \, g$ reduces to $10 \, g$ in $4 \, mi

Vedclass · Accepted Answer

In $12 \, minutes$

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(B) The half-life $T_{1/2}$ is the time taken for a substance to reduce to half its initial mass.Given that $20 \, g$ reduces to $10 \, g$ in $4 \, minutes$, the half-life $T_{1/2} = 4 \, minutes$.We use the radioactive decay formula: $M = M_0 \left( \frac{1}{2} \right)^{t / T_{1/2}}$, where $M$ is the final mass, $M_0$ is the initial mass, and $t$ is the time elapsed.Substituting the values $M = 10 \, g$, $M_0 = 80 \, g$, and $T_{1/2} = 4 \, minutes$:$10 = 80 \left( \frac{1}{2} \right)^{t / 4}$$\frac{10}{80} = \left( \frac{1}{2} \right)^{t / 4}$$\frac{1}{8} = \left( \frac{1}{2} \right)^{t / 4}$Since $\frac{1}{8} = \left( \frac{1}{2} \right)^3$, we have:$\left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{t / 4}$Equating the exponents: $3 = \frac{t}{4}$$t = 12 \, minutes$.

If $20 \, g$ of a radioactive substance reduces to $10 \, g$ due to radioactive decay in $4 \, minutes$, then in what time will $80 \, g$ of the same substance reduce to $10 \, g$?

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