(B) Let $N$ be the amount of radioactive isotope $X$ remaining and $N_0$ be the initial amount.The amount of stable element $Y$ formed is $N_0 - N$.Gi

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(B) Let $N$ be the amount of radioactive isotope $X$ remaining and $N_0$ be the initial amount.The amount of stable element $Y$ formed is $N_0 - N$.Given the ratio $\frac{N}{N_0 - N} = \frac{1}{15}$.This implies $\frac{N_0 - N}{N} = 15$,so $\frac{N_0}{N} - 1 = 15$,which gives $\frac{N_0}{N} = 16$.We know the radioactive decay law: $N = N_0 (\frac{1}{2})^{t/T_{1/2}}$,where $T_{1/2} = 50$ years.Thus,$\frac{N_0}{N} = 2^{t/T_{1/2}} = 16$.Since $16 = 2^4$,we have $\frac{t}{T_{1/2}} = 4$.Therefore,$t = 4 \times 50 = 200$ years.

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

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The half-life of a radioactive isotope $X$ is $50$ years. It decays to another element $Y$ which is stable. The two elements $X$ and $Y$ were found to be in the ratio of $1 : 15$ in a sample of a given rock. The age of the rock was estimated to be:

A
$150$
B
$200$
C
$250$
D
$100$

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Class 12

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Chapter

Nuclei

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

Samples of two radioactive nuclides,$X$ and $Y$,each have equal activity $A_0$ at time $t = 0$. $X$ has a half-life of $24$ years and $Y$ has a half-life of $16$ years. The samples are mixed together. What will be the total activity of the mixture at $t = 48$ years?

Medium

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If $N_t = N_o e^{-\lambda t}$,then the number of nuclei decaying between time $t_1$ and $t_2$ $(t_2 > t_1)$ is given by:

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$A$ radioactive sample decays by $\beta$-emission. In the first $2 \ s$,$n$ $\beta$-particles are emitted,and in the next $2 \ s$,$0.25n$ $\beta$-particles are emitted. The half-life of the radioactive nuclei is ...... $s$.

Medium

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In a nuclear reactor,the activity of a radioactive substance is $2000 / s$. If the mean life of the products is $50 \text{ minutes}$,then in the steady power generation,the number of radionuclides is:

EasyAP EAMCET 2019

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$A$ nucleus has a half-life of $30 \; min$. At $3 \; PM$, its decay rate was measured as $120000 \; cps$. What is the decay rate in $cps$ at $5 \; PM$?

MediumKVPY 2010

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The half-life of a radioactive isotope $X$ is $50$ years. It decays to another element $Y$ which is stable. The two elements $X$ and $Y$ were found to be in the ratio of $1 : 15$ in a sample of a given rock. The age of the rock was estimated to be:

Similar Questions

Samples of two radioactive nuclides,$X$ and $Y$,each have equal activity $A_0$ at time $t = 0$. $X$ has a half-life of $24$ years and $Y$ has a half-life of $16$ years. The samples are mixed together. What will be the total activity of the mixture at $t = 48$ years?

If $N_t = N_o e^{-\lambda t}$,then the number of nuclei decaying between time $t_1$ and $t_2$ $(t_2 > t_1)$ is given by:

$A$ radioactive sample decays by $\beta$-emission. In the first $2 \ s$,$n$ $\beta$-particles are emitted,and in the next $2 \ s$,$0.25n$ $\beta$-particles are emitted. The half-life of the radioactive nuclei is ...... $s$.

In a nuclear reactor,the activity of a radioactive substance is $2000 / s$. If the mean life of the products is $50 \text{ minutes}$,then in the steady power generation,the number of radionuclides is:

$A$ nucleus has a half-life of $30 \; min$. At $3 \; PM$, its decay rate was measured as $120000 \; cps$. What is the decay rate in $cps$ at $5 \; PM$?

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