Half life of radium is 1620 years and its atomic weight is 226, k,gm per kilomol. number of atoms th

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13.Nuclei

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The half life of radium is $1620$ years and its atomic weight is $226\, k\,gm$ per kilomol. The number of atoms that will decay from its $1\, gm$ sample per second will be

(Avogadro's number $N = 6.02 \times {10^{26}}$atom/kilomol)

$3.61 \times {10^{10}}$

$3.6 \times {10^{12}}$

$3.11 \times {10^{15}}$

$31.1 \times {10^{15}}$

Solution

(a) $\frac{{dN}}{{dt}} = \lambda N;$

$\lambda = \frac{{0.6931}}{{{t_{12}}}} = \frac{{0.6931}}{{1620 \times 365 \times 24 \times 60 \times 60}}$

$N = \frac{{6.023 \times {{10}^{23}}}}{{226}}$

$\therefore \frac{{dN}}{{dt}} = \frac{{0.6931 \times 6.023 \times {{10}^{23}}}}{{1620 \times 365 \times 24 \times 60 \times 60 \times 226}} = 3.61 \times 10^{10}$

Standard 12

Physics

Radon $({R_n})$ decays into Polonium (${P_0}$) by emitting an $\alpha – $ particle with half-life of $4\, days$. A sample contains $6.4 \times {10^{10}}$ atoms of $R_n$. After $12\, days$, the number of atoms of ${R_n}$ left in the sample will be

easy

The phenomenon of radioactivity is

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Two radioactive nuclei $P$ and $Q,$ in a given sample decay into a stable nucleus $R.$ At time $t = 0,$ number of $P$ species are $4\,\, N_0$ and that of $Q$ are $N_0$. Half-life of $P$ (for conversion to $R$) is $1$ minute where as that of $Q$ is $2$ minutes. Initially there are no nuclei of $R$ present in the sample. When number of nuclei of $P$ and $Q$ are equal, the number of nuclei of $R$ present in the sample would be

hard

(AIPMT-2011)

The graph between number of decayed atoms $N'$ of a radioactive element and time $t$ is

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The count rate for $10\ g$ of radioactive material was measured at different times and this has been shown in the graph with scale given. The half life of the material and the total count in the first half value period respectively are