Two radioactive elements $R$ and $S$ disintegrate as:
$R \rightarrow P + \alpha; \lambda_R = 4.5 \times 10^{-3} \, \text{years}^{-1}$
$S \rightarrow P + \beta; \lambda_S = 3 \times 10^{-3} \, \text{years}^{-1}$
Starting with the number of atoms of $R$ and $S$ in the ratio of $2:1$,what will be this ratio after the lapse of three half-lives of $R$?

  • A
    $3:2$
  • B
    $1:3$
  • C
    $1:1$
  • D
    $2:1$

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Similar Questions

If the measurement errors in all the independent quantities are known,then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example,consider the relation $z = x / y$. If the errors in $x, y$ and $z$ are $\Delta x, \Delta y$ and $\Delta z$,respectively,then $z \pm \Delta z = \frac{x \pm \Delta x}{y \pm \Delta y} = \frac{x}{y}(1 \pm \frac{\Delta x}{x})(1 \pm \frac{\Delta y}{y})^{-1}$. The series expansion for $(1 \pm \frac{\Delta y}{y})^{-1}$,to first power in $\Delta y / y$,is $1 \mp(\Delta y / y)$. The relative errors in independent variables are always added. So the error in $z$ will be $\Delta z = z(\frac{\Delta x}{x} + \frac{\Delta y}{y})$. The above derivation makes the assumption that $\Delta x / x \ll 1, \Delta y / y \ll 1$. Therefore,the higher powers of these quantities are neglected.
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$(2)$ In an experiment,the initial number of radioactive nuclei is $3000$. It is found that $1000 \pm 40$ nuclei decayed in the first $1.0 \ s$. For $|x| \ll 1$,$\ln(1+x) \approx x$ up to the first power in $x$. The error $\Delta \lambda$,in the determination of the decay constant $\lambda$,in $s^{-1}$,is:

$A$ sample initially contains only $U-238$ isotope of uranium. With time,some of the $U-238$ radioactively decays into $Pb-206$ while the rest of it remains undisintegrated. When the age of the sample is $P \times 10^8$ years,the ratio of the mass of $Pb-206$ to that of $U-238$ in the sample is found to be $7$. The value of $P$ is. . . . . . [Given: Half-life of $U-238$ is $4.5 \times 10^9$ years; $\log_e 2 = 0.693$]

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$A$ solution containing active cobalt ${}_{27}^{60}Co$ having activity of $0.8\,\mu Ci$ and decay constant $\lambda$ is injected into an animal's body. If $1\,cm^3$ of blood is drawn from the animal's body after $10\,hrs$ of injection,the activity found is $300\,decays$ per minute. What is the total volume of blood in the animal's body in litres? (Given: $1\,Ci = 3.7 \times 10^{10}$ decays per second and at $t = 10\,hrs$,$e^{-\lambda t} = 0.84$)

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