The half-life of a radioactive substance is $30$ minutes. The time (in minutes) taken between $40\%$ decay and $85\%$ decay of the same radioactive substance is

Consider an initially pure $M \text{ g}$ sample of an isotope $X$ with mass number $A$,which has a half-life of $T \text{ hours}$. What is its initial decay rate? ($N_A$ = Avogadro number)

The counting rate observed from a radioactive source at $t = 0$ was $1600 \ counts \ s^{-1}$,and at $t = 8 \ s$,it was $100 \ counts \ s^{-1}$. The counting rate observed as $counts \ s^{-1}$ at $t = 6 \ s$ will be

$C^{14}$ has a half-life of $5700$ years. At the end of $11400$ years,the actual amount left is:

Activity of a radioactive sample is $R_1$ at a time $t_1$ and $R_2$ at a time $t_2$. Its half-life period is $T$. The number of atoms that have disintegrated in the time interval $(t_2 - t_1)$ is equal to $\frac{n(R_1 - R_2)T}{\ln 4}$. Then '$n$' is equal to

$A$ $280\, \text{day}$ 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? ......... $\text{dps}$