Activity of a radioactive substance is $R_1$ at time $t_1$ and $R_2$ at time $t_2(t_2 > t_1).$ Then the ratio $\frac{R_2}{R_1}$ is :
$\frac{t_2}{t_1}$
${e^{ - \lambda ({t_1} + {t_2})}}$
$e\left( {\frac{{{t_1} - {t_2}}}{\lambda }} \right)$
${e^{ \lambda ({t_1} + {t_2})}}$
A radioactive sample $\mathrm{S} 1$ having an activity $5 \mu \mathrm{Ci}$ has twice the number of nuclei as another sample $\mathrm{S} 2$ which has an activity of $10 \mu \mathrm{Ci}$. The half lives of $\mathrm{S} 1$ and $\mathrm{S} 2$ can be
The nuclide $^{131}I$ is radioactive, with a half-life of $8.04$ days. At noon on January $1$, the activity of a certain sample is $60089$. The activity at noon on January $24$ will be
For a certain radioactive process the graph between $In\, {R}$ and ${t}\,({sec})$ is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately $....\,{sec}.$
The half-life of ${ }^{198} {Au}$ is $3 \,days.$ If atomic weight of ${ }^{198} {Au}$ is $198\, {g} / {mol}$ then the activity of $2 \,{mg}$ of ${ }^{198} {Au}$ is ..... $\times 10^{12}\,disintegration/second$
A freshly prepared sample of a radioisotope of half-life $1386 \ s$ has activity $10^3$ disintegrations per second. Given that In $2=0.693$, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first $80 \ s$ after preparation of the sample is :