The half-life of a particular radioactive isotope is 6.5h. If there are initially$\mathbf{48}\mathbf{\times }{\mathbf{10}}^{\mathbf{19}}$ atoms of this isotope, how many remain at the end of 26h?

a) Half-life of the given radioactive isotope $T_{1/2}=6.5\mathrm{h}$,

b) Initially, the number of atoms of the isotope,$N_0=48\times {10}^{19}$atoms

c) The time of decay t = 26h,

Consider the radioactive decay constant or the disintegration constant represents the fraction of radioactive atoms that disintegrates in a unit of time. The given relation gives the number of atoms remaining at the end of the decay after a given time of the given isotope.

Formulae:

The disintegration constant is as follows:

$\mathrm{\lambda }=\frac{\mathrm{In}2}{{\mathrm{T}}_{{\displaystyle \frac{1}{2}}}}$ …… (i)

Here,$T_{\frac{1}{2}}$is the half-life of the substance.

The undecayed sample remaining after a given time as follows:

$N=N_0{e}^{-\lambda t}$ ….. (ii)

After substituting equation (i) in equation (ii) and then using the given data, the remaining undecayed atoms of the given radioactive isotope as follows:

$N=N_0{e}^{\frac{\mathrm{In}2}{{\mathrm{T}}_{{\displaystyle \frac{1}{2}}}}\left(t\right)}$

Solve further as:

$$\begin{gathered} N=\left(48\times {10}^{19}\right)e^{\frac{\mathrm{In}2}{6.5\mathrm{h}}\left(26\mathrm{h}\right)} \\ =\left(48\times {10}^{19}\right){\mathrm{e}}^{-4}\mathrm{In}2 \\ =3\times {10}^{19} \end{gathered}$$

Hence, the number of remaining atoms is $3\times {10}^{19}$.