Substance $A$ has an atomic mass number of $16$ and a half-life of $1$ day. Another substance $B$ has an atomic mass number of $32$ and a half-life of $0.5$ day. If both $A$ and $B$ start undergoing radioactivity simultaneously with an initial mass of $320 \, g$ each,how many total atoms of $A$ and $B$ combined would be left after $2$ days? (Answer in $......... \times 10^{24}$)

$A$ small quantity of solution containing $Na^{24}$ radionuclide of activity $1 \, \mu Ci$ is injected into the blood of a person. $A$ sample of the blood of volume $1 \, cm^3$ taken after $5 \, hours$ shows an activity of $296$ disintegrations per minute. What will be the total volume of the blood in the body of the person? Assume that the radioactive solution mixes uniformly in the blood of the person: ............ $L$ (Take $1 \, Ci = 3.7 \times 10^{10}$ disintegrations per second and $e^{-\lambda t} = 0.7927$; where $\lambda$ is the disintegration constant).

Consider a radioactive nuclide which follows the decay rate given by $A(t) = A_0 2^{-(t/t_0)}$, where $A(t)$ is the fraction of radioactive material remaining after time $t$ from the initial $A_0$ at zero time. Let $A_1$ be the fraction of original activity which remains after $120 \ h$. Likewise, $A_2$ is the fraction of original activity remaining after $200 \ h$. If $A_1/A_2 = 16$, then the half-life $(t_0)$ will be: (in $h$)

If the half-life of a radioactive atom is $2.3 \, days$,then its decay constant would be

If one starts with $1 \, \text{curie}$ of a radioactive substance $(T_{1/2} = 12 \, \text{hrs})$, the activity left after a period of $1 \, \text{week}$ will be about:

$A$ radioactive nucleus $A$ has a single decay mode with half-life $\tau_A$. Another radioactive nucleus $B$ has two decay modes $1$ and $2$. If decay mode $2$ were absent,the half-life of $B$ would have been $\tau_A / 2$. If decay mode $1$ were absent,the half-life of $B$ would have been $3 \tau_A$. If the actual half-life of $B$ is $\tau_B$,then the ratio $\tau_B / \tau_A$ is