A) \[~45.6\text{ }\times \text{ }{{10}^{23}}\text{ atom}\]
B) \[4.56\text{ }\times \text{ }{{10}^{23}}\text{ atom}\]
C) \[4.56\text{ }\times \text{ }{{10}^{21}}\text{ atom}\]
D) \[4.56\text{ }\times \,\,{{10}^{20}}\text{ atom}\]
Correct Answer: B
Solution :
\[\text{k = }\frac{0.693}{{{t}_{1/2}}}=\frac{0.693}{10}h{{r}^{-1}}\] \[k=\frac{2.303}{4}\log \frac{1\times 6.023\times {{10}^{23}}}{N}\] \[(\because \,1\,g\,\text{atom}=6.023\times {{10}^{23}}\,\text{atom})\] \[\Rightarrow \] \[\frac{0.693}{10}=\frac{2.303}{4}\log \frac{6.023\times {{10}^{23}}}{N}\] or \[\log \frac{6.023\times {{10}^{23}}}{N}=0.12036\] \[\frac{6.023\times {{10}^{23}}}{N}=1.319\] \[\therefore \] \[N=4.56\times {{10}^{23}}\text{atom}\text{.}\]
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