Consider an initially pure3.4g sample of${}^{\mathbf{67}}\mathit{G}\mathit{a}$, an isotope that has a half-life of 78h. (a) What is its initial decay rate? (b) What is its decay rate 48hlater?

1. Mass of the sample ${}^{67}\mathrm{Ga}$, m = 3.4 g
2. Half-life of the isotope ${}^{67}\mathrm{Ga}$,$T_{1/2}=78\mathrm{h}$
3. Time of decay, t = 48h
4. Molar mass of the isotope ${}^{67}\mathrm{Ga}$,role="math" localid="1661923366531" $A=67\frac{9}{\mathrm{mo}l}$

The radioactive decay is due to the loss of the elementary particles from an unstable nucleus to convert them into a more stable one. 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. Now, for each case of decay, calculate the total number of atoms present in the given mass of the Gallium sample using the Avogadro number.

Formulae:

The rate of decay is as follows:

$R=\lambda N$ ….. (i)

Here,$\lambda$is the disintegration constant, and Nis the number of undecayed nuclei.

The disintegration constant is as follows:

$1=\frac{\ln 2}{T_{1/2}}$ …… (ii)

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

The undecayed sample remaining after a given time is as follows:

$N=N_0{e}^{-\lambda t}$ ….. (iii)

The number of undecayed nuclei in a given mass of a substance is as follows:

$N=\frac{m}{A}{N}_A$ ….. (iv)

Here,$N_A=6.022\times {10}^{23}\frac{nuclei}{mol}$

Consider the disintegration constant of the sample using the data in equation (ii) as follows:

$\begin{array}{l}\lambda =\frac{\ln 2}{78h}\\ =8.89\times {10}^{-3}{h}^{-1}\end{array}$

Consider the number of nuclei present in the given mass of the isotope${}^{67}\mathrm{Ga}$ at the initial state can be given using equation (iv) as follows:

$$\begin{gathered} N_0=\frac{\left(3.4g\right)}{67g/mol}\left(6.022\times {10}^{23}\frac{nuclei}{mol}\right) \\ =3.055\times {10}^{22}nuclei \end{gathered}$$

Thus, the rate of initial decay can be given using equation (i) as follows:

$\begin{array}{l}{R}_0=\left(8.89\times {10}^{-3}{h}^{-1}\right)\left(3.055\times {10}^{22}nuclei\right)\\ =2.71\times {10}^{20}{h}^{-1}\\ =7.53\times {10}^{16}{s}^{-1}\end{array}$

Hence, the value of the decay rate is $7.53\times {10}^{16}{s}^{-1}$.

Multiplying the disintegration constant to both the sides of the equation (iii) and comparing it with equation (i), get the rate of undecayed nuclei of the isotope${}^{67}\mathrm{Ga}$ at the end of as follows:

$R=R_0{e}^{-\lambda t}$

Solve further as:

role="math" localid="1661924058017" $\begin{array}{lll}R=7.53\times {10}^{16}{s}^{-1}{e}^{-\left(8.89\times {10}^{-3}{h}^{-1}\right)\left(48h\right)}& & \\ =7.53\times {10}^{16}{s}^{-1}{e}^{-0.427}& & \\ =4.91\times {10}^{16}{s}^{-1}& & \end{array}$

Hence, the value of decay rate is $4.91\times {10}^{16}{s}^{-1}$.