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:

$A$ sample of a radioactive element contains $4 \times 10^{16}$ active nuclei. If the half-life of the element is $10$ days,then the number of decayed nuclei after $30$ days is ........ $\times 10^{16}$.

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.
$(1)$ Consider the ratio $r = \frac{(1-a)}{(1+a)}$ to be determined by measuring a dimensionless quantity $a$. If the error in the measurement of $a$ is $\Delta a$ $(\Delta a / a \ll 1)$,then what is the error $\Delta r$?
$(2)$ 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:

The half-life of $Ra^{226}$ is $1620$ years. Calculate the number of atoms that decay in one second in $1 \ g$ of radium (Avogadro number $= 6.023 \times 10^{23}$).

After two hours, one-sixteenth of the starting amount of a certain radioactive isotope remained undecayed. The half-life of the isotope is:

$A$ radioactive sample at any instant has its disintegration rate $5000$ disintegrations per minute. After $5$ minutes,the rate is $1250$ disintegrations per minute. Then,the decay constant (per minute) is (in $, \ln 2$)