Assertion : If the half-life of a radioactive | vedclass

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(D) The half-life $(T_{1/2})$ of the radioactive substance is $40 \ days$. The amount of substance remaining after time $t$ is given by $N = N_0 \left

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If the Assertion is incorrect but the Reason is correct.

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(D) The half-life $(T_{1/2})$ of the radioactive substance is $40 \ days$. The amount of substance remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} ight)^{t/T_{1/2}}$.For $t = 20 \ days$,the number of half-lives is $n = \frac{20}{40} = 0.5$.The amount remaining is $N = N_0 \left( \frac{1}{2} ight)^{0.5} = N_0 \left( \frac{1}{\sqrt{2}} ight) \approx 0.707 N_0$.The amount decayed is $N_0 - N = N_0 - 0.707 N_0 = 0.293 N_0$,which is $29.3\%$.Since $29.3\% 
eq 25\%$,the Assertion is incorrect.The formula provided in the Reason is the standard law of radioactive decay,which is correct.Therefore,the Assertion is incorrect but the Reason is correct.

Assertion : If the half-life of a radioactive substance is $40 \ days$,then $25\%$ of the substance decays in $20 \ days$.
Reason : $N = N_0 \left( \frac{1}{2} \right)^n$,where $n = \frac{\text{time elapsed}}{\text{half-life period}}$.

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Assertion : If the half-life of a radioactive substance is $40 \ days$,then $25\%$ of the substance decays in $20 \ days$.Reason : $N = N_0 \left( \frac{1}{2} \right)^n$,where $n = \frac{\text{time elapsed}}{\text{half-life period}}$.