Question

Half life of ${}_{82}^{214}X$ is 26.8 minutes.Mass of one curie of X is

• A. $3.04\times {10}^{-11}kg$
• B. $3.05\times {10}^{23}kg$
• C. $8.61\times {10}^{10}kg$
• D. $8.06\times {10}^{-8}kg$

Answer 1

Answer: (A)

$3.04\times {10}^{-11}kg$

$\begin{array}{c}\text{Given,}\\ \text{half life}\left(T_{1/2}\right){\text{of}}_{82}^{214}X=26.8\text{minute}\end{array}$

$\begin{array}{c}\text{Now, we know}\\ \begin{array}{cc}\lambda & =\frac{0.693}{T_{1/2}}\\ \lambda & =\frac{0.693}{26.8\times 60}\\ \lambda & =4.32\times {10}^{-4}{s}^{-1}\end{array}\end{array}$

$\begin{array}{c}\text{Now, we know}\\ 1\text{curie}=3.7\times {10}^{10}\text{dps}\end{array}$

$\begin{array}{c}\begin{array}{c}\text{Let number of atoms in}1\text{curie}=N\\ \frac{dN}{dt}=\lambda N\\ \frac{dN}{dt}=3.7\times {10}^{10}dps\\ 3\cdot 7\times {10}^{10}=4.32\times {10}^{-4}\times N\\ N=\frac{3.7\times {10}^{10}}{4.32\times {10}^{-4}}\\ N=0.856\times {10}^{14}\end{array}\end{array}$

$\begin{array}{cc}\text{mass of}1\text{curie}& =\frac{\mu }{\text{avogadro}{N}_0}\times \text{molar mass}\\ & =\frac{0.856\times {10}^{14}}{6.022\times {10}^{23}}\times 214\\ m& =3.04\times {10}^{-11}kg\end{array}$