Question

A$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{g}$sample of samarium emits alpha particles at a rate of 120 particles/s. The responsible isotope is${}^{\mathbf{147}}\mathbf{Sm}$whose natural abundance in bulk samarium is 15.0%. Calculate the half-life.

Short answer

The half-life of the samarium isotope${}^{147}\mathrm{Sm}$ is $1.12\times {10}^{11}y$.

Step-by-step solution

The given data

a) The mass of the samarium sample, m=1.00g

b) Molar mass of the ${}^{147}\mathrm{Sm}$ isotope, A=147g/mol

c) Rate of decay of the alpha particles, R=120particles/s

d) The natural abundance of the isotope${}^{147}\mathrm{Sm}$ is 15.0%.

Understanding the concept of natural abundance of isotope  

A substance is always present in the bulk form in the natural consisting of its isotopes. Here, we are provided with the samarium sample, whose is present in only of nature. Thus, in the total given amount, only a portion of the number of nuclei of the samarium sample is the required nuclei number for the given isotope. Thus, a smaller portion of the sample implies the low stable nature of the isotope.

Formula:

The rate of decay,$\mathbf{R}\mathbf{=}\mathbf{\lambda N}$ (i)

Where,$\mathit{\lambda }$is the disintegration constant

NIs the number of undecayed nuclei

The disintegration constant,$\mathbf{\lambda }\mathbf{=}\frac{\mathbf{In}\mathbf{}\mathbf{2}}{{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}}$ (ii)

Where,${\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}$is the half-life of the substance.

The number of nuclei in a given mass of a substance,

$\mathbf{N}\mathbf{=}\frac{\mathbf{m}}{\mathbf{A}}{\mathbf{N}}_{\mathbf{A}}\mathbf{,}{\mathbf{whereN}}_{\mathbf{A}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}\mathbf{}}\mathbf{nuclei}\mathbf{/}\mathbf{mol}$ (iii)

Calculation of the half-life

As per the given problem, only 15% of the sample is the given isotope ${}^{147}\mathrm{Sm}$. Thus, the number of nuclei of the${}^{147}\mathrm{Sm}$ isotope can be given using equation (iii) as follows:

$$\begin{gathered} N_{{}^{147}\mathrm{Sm}}=\frac{0.15\left(1.00\mathrm{g}\right)}{\left(147\mathrm{g}/\mathrm{mol}\right)}\left(6.022\times {10}^{23}\mathrm{nuclei}/\mathrm{mol}\right) \\ =6.143\times {10}^{20}\mathrm{nuclei} \end{gathered}$$

Now, substituting equation (ii) in equation (i) and using the above nuclei number, we get that the half-life of the isotope is given as follows:

role="math" localid="1661600408120" $$\begin{gathered} R=\left(\frac{In2}{T_{1/2}}\right)N \\ {T}_{1/2}=\left(\frac{In2}{R}\right)N \\ =\left(\frac{In2}{120/s}\right)6.143\times {10}^{20} \\ =3.55\times {10}^{18}s \\ =1.12\times {10}^{11}y \end{gathered}$$

Hence, the value of the half-life is $1.12\times {10}^{11}y$.