Question

The half-life of a radioactive nuclide is $20h$. It is found that the fraction $\left(1/x\right)$ of original activity remains after $40$ hours? What is the value of $x?$

Answer 1

Half life, $t_{1/2}=\frac{0.693}{\lambda }$ where $\lambda =$ decay constant.
Using, $N=N_0{e}^{-\lambda t}$
Here, $N_0=1,N=\frac{1}{x}$ so , $\frac{N_0}{N}=e^{\lambda t}$

$\frac{1}{1/x}=e^{\frac{t_{1/2}}{0.693}t}=e^{\frac{20}{0.693}\left(40\right)}$

$x=3.99\sim 4$