A radioactive material has a half-life of $8$ years. The activity of the material will decrease to about $\frac{1}{8}$ th of its original value in :

256

years

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128

years

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64

years

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24

years

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Solution

The correct option is D $24$ years
$R = R_{0} {(\frac{1}{2})}^{n}$
$\frac{R}{R_{0}} = {(\frac{1}{2})}^{t / T_{1 / 2}}$
$(\frac{1}{8}) = {(\frac{1}{2})}^{t / T_{1 / 2}}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{t / T_{1 / 2}}$
$\frac{t}{T_{1 / 2}} = 3$
$t = 3 T_{1 / 2} = 3 \times 8 = 24$ years

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A radioactive material has a half-life of 8 years. The activity of the material will decrease to about 18th of its original value in :

The correct option is D 24 yearsR=R0(12)nRR0=(12)t/T1/2(18)=(12)t/T1/2(12)3=(12)t/T1/2tT1/2=3t=3T1/2=3×8=24 years

A radioactive material has a half-life of $8$ years. The activity of the material will decrease to about $\frac{1}{8}$ th of its original value in :

The correct option is D $24$ years
$R = R_{0} {(\frac{1}{2})}^{n}$
$\frac{R}{R_{0}} = {(\frac{1}{2})}^{t / T_{1 / 2}}$
$(\frac{1}{8}) = {(\frac{1}{2})}^{t / T_{1 / 2}}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{t / T_{1 / 2}}$
$\frac{t}{T_{1 / 2}} = 3$
$t = 3 T_{1 / 2} = 3 \times 8 = 24$ years