After two hours, one- sixteenth of the starting amount of a certain radioactive isotope remained undecayed. The half life of the isotope is
$15\, minutes$
$30 \,minutes$
$45 \,minutes$
$1 \,hour$
In a radioactive sample, ${ }_{10}^a K$ nuclei either decay into stable ${ }_{20}^{* 0} Ca$ nuclei with decay constant $4.5 \times 10^{-10}$ per year or into stable ${ }_{18}^{40}$ Ar muclei with decay constant $0.5 \times 10^{-10}$ per year. Given that in this sample all the stable ${ }_{20}^{\infty 0} Ca$ and ${ }_{15}^{20} Ar$ nuclei are produced by the ${ }_{19}^{* 0} K$ muclei only. In time $t \times 10^{\circ}$ years, if the ratio of the sum of stable ${ }_{30}^{40} Ca$ and ${ }_{15} \operatorname{An}$ nuclei to the radioactive ${ }_{19} K$ muclei is $99$ , the ralue of $t$ will be : [Given $\ln 10=2.3]$
The relation between $\lambda $ and $({T_{1/2}})$ is (${T_{1/2}}=$ half life, $\lambda=$ decay constant)
The mean lives of a radioactive sample are $30$ years and $60$ years for $\alpha$-emission and $\beta $ -emission respectively. If the sample decays both by $\alpha$- emission and $\beta $-emission simultaneously, the time after which, only one-fourth of the sample remain is :- ........... $years$
The decay constant of a radioactive element is $0.01$ per second. Its half life period is .......$sec$
${ }^{131} I$ is an isotope of Iodine that $\beta$ decays to an isotope of Xenon with a half-life of $8$ days. A small amount of a serum labelled with ${ }^{131} \mathrm{I}$ is injected into the blood of a person. The activity of the amount of ${ }^{131} \mathrm{I}$ injected was $2.4 \times 10^5$ Becquerel $(\mathrm{Bq})$. It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After $11.5$ hours, $2.5 \mathrm{ml}$ of blood is drawn from the person's body, and gives an activity of $115 \mathrm{~Bq}$. The total volume of blood in the person's body, in liters is approximately (you may use $\mathrm{e}^{\mathrm{x}} \approx 1+\mathrm{x}$ for $|\mathrm{x}| \ll 1$ and $\ln 2 \approx 0.7$ ).