(D) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}}$,where $t = 12\, days$ and $T_{1/2} = 4\, days$.Thus,$n = \frac{12}{4} = 3$.The nu

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(D) The number of half-lives $n$ is given by $n = \frac{t}{T_{1/2}}$,where $t = 12\, days$ and $T_{1/2} = 4\, days$.Thus,$n = \frac{12}{4} = 3$.The number of atoms remaining $N$ is given by the formula $N = N_0 \times (\frac{1}{2})^n$,where $N_0 = 6.4 \times 10^{10}$.Substituting the values,we get $N = 6.4 \times 10^{10} \times (\frac{1}{2})^3$.$N = 6.4 \times 10^{10} \times \frac{1}{8} = 0.8 \times 10^{10}$ atoms.

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

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Radon $({Rn})$ decays into Polonium $({Po})$ by emitting an $\alpha$-particle with a half-life of $4\, days$. $A$ sample contains $6.4 \times 10^{10}$ atoms of $Rn$. After $12\, days$,the number of atoms of $Rn$ left in the sample will be:

A
$3.2 \times 10^{10}$
B
$0.53 \times 10^{10}$
C
$2.1 \times 10^{10}$
D
$0.8 \times 10^{10}$

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Class 12 Questions

Class 12

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Nuclei

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

The activity of a radioactive sample is measured as $N_0$ counts per minute at $t = 0$ and $N_0/e$ counts per minute at $t = 5 \, minutes$. The time (in $minutes$) at which the activity reduces to half its value is

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Two radioactive isotopes $P$ and $Q$ have half-lives $10 \text{ minutes}$ and $15 \text{ minutes}$ respectively. Freshly prepared samples of each isotope initially contain the same number of atoms. After $30 \text{ minutes}$,the ratio $\frac{\text{number of atoms of } P}{\text{number of atoms of } Q}$ will be

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$A$ mixture consists of two radioactive materials $A_1$ and $A_2$ with half-lives of $20 \, s$ and $10 \, s$ respectively. Initially,the mixture has $40 \, g$ of $A_1$ and $160 \, g$ of $A_2$. The amount of the two in the mixture will become equal after ......... $s$.

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$N$ atoms of a radioactive element emit $n$ alpha particles per second. The half-life of the element is:

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$A$ fraction $f_1$ of a radioactive sample decays in one mean life,and a fraction $f_2$ decays in one half-life.

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Radon $({Rn})$ decays into Polonium $({Po})$ by emitting an $\alpha$-particle with a half-life of $4\, days$. $A$ sample contains $6.4 \times 10^{10}$ atoms of $Rn$. After $12\, days$,the number of atoms of $Rn$ left in the sample will be:

Similar Questions

The activity of a radioactive sample is measured as $N_0$ counts per minute at $t = 0$ and $N_0/e$ counts per minute at $t = 5 \, minutes$. The time (in $minutes$) at which the activity reduces to half its value is

$A$ mixture consists of two radioactive materials $A_1$ and $A_2$ with half-lives of $20 \, s$ and $10 \, s$ respectively. Initially,the mixture has $40 \, g$ of $A_1$ and $160 \, g$ of $A_2$. The amount of the two in the mixture will become equal after ......... $s$.

$N$ atoms of a radioactive element emit $n$ alpha particles per second. The half-life of the element is:

$A$ fraction $f_1$ of a radioactive sample decays in one mean life,and a fraction $f_2$ decays in one half-life.

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