$A$ certain radioactive substance reduces to $25\%$ of its initial value in $16\ days$. Its half-life is .......... $days$.

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 $h$)

$A$ radioactive element has a rate of disintegration of $9000$ disintegrations per minute at a particular instant. After $2$ minutes, it becomes $3000$ disintegrations per minute. The decay constant per minute is:

$A$ $280\, \text{day}$ old radioactive substance shows an activity of $6000\, \text{dps}$. $140\, \text{days}$ later, its activity becomes $3000\, \text{dps}$. What was its initial activity? ......... $\text{dps}$

If the mass of a radioactive sample is doubled,the activity of the sample and the disintegration constant of the sample are respectively

The half-life of a neutron is $693 \ s$. What fraction of neutrons will decay when a beam of neutrons, having kinetic energy of $0.084 \ eV$, travels a distance of $1 \ km$? (mass of neutron $= 1.68 \times 10^{-27} \ kg$, and $\ln 2 = 0.693$)