(C) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}}$.For substance $A$,$n_A = \frac{80}{20} = 4$.For substance $B$,$n_B = \frac{80}{40

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(C) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}}$.For substance $A$,$n_A = \frac{80}{20} = 4$.For substance $B$,$n_B = \frac{80}{40} = 2$.The number of remaining nuclei is given by $N = N_0 \left( \frac{1}{2} \right)^n$.Since initial numbers $N_0$ are equal,the ratio is $\frac{N_A}{N_B} = \frac{N_0 (1/2)^{n_A}}{N_0 (1/2)^{n_B}} = \frac{(1/2)^4}{(1/2)^2} = \frac{2^2}{2^4} = \frac{4}{16} = \frac{1}{4}$.Thus,the ratio is $1 : 4$.

Class 12 Physics Nuclei Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

Medium

Half-lives of two radioactive substances $A$ and $B$ are $20 \text{ minutes}$ and $40 \text{ minutes}$ respectively. Initially,the samples of $A$ and $B$ have an equal number of nuclei. After $80 \text{ minutes}$,the ratio of the remaining number of $A$ and $B$ nuclei is:

AIPMT 1998 All AIPMT

A
$1 : 16$
B
$4 : 1$
C
$1 : 4$
D
$1 : 1$

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Nuclei

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Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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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 $n = \frac{\text{time elapsed}}{\text{half-life period}}$.

MediumAIIMS 1998

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The atomic weight of a radioactive element is $M_w$ grams. The radioactivity of $m$ grams of this element is ......... ($N_A$ is Avogadro's number,$\lambda$ is the decay constant).

Easy

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$A$ radioactive material has an initial amount $16 \, gm$. After $120 \, days$ it reduces to $1 \, gm$. The half-life of the radioactive material is .......... $days$.

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The specific activity of $Ra^{226}$ is $1 \, Ci/g$. The activity of $1 \, \mu g$ of $Ra^{226}$ will be:

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$A$ radio isotope has a half-life of $75\, \text{years}$. The fraction of the atoms of this material that would decay in $150\, \text{years}$ will be...........$\%$

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Half-lives of two radioactive substances $A$ and $B$ are $20 \text{ minutes}$ and $40 \text{ minutes}$ respectively. Initially,the samples of $A$ and $B$ have an equal number of nuclei. After $80 \text{ minutes}$,the ratio of the remaining number of $A$ and $B$ nuclei is:

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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 $n = \frac{\text{time elapsed}}{\text{half-life period}}$.

The atomic weight of a radioactive element is $M_w$ grams. The radioactivity of $m$ grams of this element is ......... ($N_A$ is Avogadro's number,$\lambda$ is the decay constant).

$A$ radioactive material has an initial amount $16 \, gm$. After $120 \, days$ it reduces to $1 \, gm$. The half-life of the radioactive material is .......... $days$.

The specific activity of $Ra^{226}$ is $1 \, Ci/g$. The activity of $1 \, \mu g$ of $Ra^{226}$ will be:

$A$ radio isotope has a half-life of $75\, \text{years}$. The fraction of the atoms of this material that would decay in $150\, \text{years}$ will be...........$\%$

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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 $n = \frac{\text{time elapsed}}{\text{half-life period}}$.