91Ó°ÊÓ

a) The half-life of radioactive isotope ${}^{239}\mathrm{Pu},{\mathrm{T}}_{1/2}=2.41×{10}^4\mathrm{y}$,

b) The mass of the lethal dose of Plutonium,$m=2×{10}^{-30}\mathrm{g}$

c) The molar mass of the Plutonium,$A=239\frac{\mathrm{g}}{\mathrm{mol}}$

One mole of a substance contains an Avogadro number of molecules or atoms or nuclei. Thus, using this concept the required value of the nuclei present in the isotope is calculated. The radioactive decay is due to the loss of the elementary particles from an unstable nucleus to convert them into a more stable one. Using the basic formula, we can get the decay rate of the plutonium substance.

Formulae:

The number of atoms in a given mass of an atom,

$$\begin{gathered} N=\frac{m}{A}{N}_A \\ \mathrm{Here},{\mathrm{N}}_{\mathrm{A}}=6.022×{10}^{23}\frac{\mathrm{nuclei}}{\mathrm{mol}} \end{gathered}$$ …… (i)

Consider the formula for the rate of decay as:

$R=\frac{N\mathrm{In}2}{T_{{\displaystyle \frac{1}{2}}}}$ …… (ii)

Using the given data in equation (i), determine the number of Pu nuclei that constitute a chemically lethal dose of 2.00 mg as follows:

$$\begin{gathered} {\mathrm{N}}_{\mathrm{Pu}}=\frac{2×{10}^{-3}\mathrm{g}}{239{\displaystyle \frac{\mathrm{g}}{\mathrm{mol}}}}\left(6.022×{10}^{23}\frac{\mathrm{nuclei}}{\mathrm{mol}}\right) \\ =5.04×{10}^{18}\mathrm{nuclei} \\ \mathrm{Hence},\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{nuclei}\mathrm{is}5.04×{10}^{18}. \end{gathered}$$

Hence, the number of nuclei is.

The amount of decay rate of the isotope is calculated using equation (ii) as follows:

$$\begin{gathered} R=\frac{\left(5.04×{10}^{18}\mathrm{nuclei}\right)\mathrm{In}2}{\left(2.41×{10}^4\mathrm{y}\right)} \\ =1.4×{10}^{14}{y}^{-1} \end{gathered}$$

Hence, the value of the decay rate is $1.4×{10}^{14}{y}^{-1}$.