Question

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?

Short answer

The amount of atoms remaining at the end of 26h is $3\times {10}^{19}$.

Step-by-step solution

Write the given data

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,

Determine the concept of undecayed atoms

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)

Calculate the remaining atoms at the end of 26h

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 $3\times {10}^{19}$.