Question

In artificial radioactivity, $ 1.414\times {{10}^{6}} $ nuclei are disintegrated into $ {{10}^{6}} $ nuclei in $ 10 \,min $ . The half-life in minutes must be:

• A. 5
• B. 20
• C. 15
• D. 30
• E. 25

Answer 1

Answer: (A)

From Rutherford-Soddy law $ N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}} $
$ n=\frac{t}{T} $ $ \therefore $
$ {{10}^{6}}=1.414\times {{10}^{6}}{{\left( \frac{1}{2} \right)}^{t/T}} $
$ \Rightarrow $ $ \frac{1000}{1414}={{\left( \frac{1}{2} \right)}^{t/T}} $
$ \Rightarrow $ $ {{\left( \frac{1}{2} \right)}^{2}}={{\left( \frac{10}{12} \right)}^{2}} $
(Approximately) $ \Rightarrow $ $ n=2 $ $ \Rightarrow $ $ n=\frac{t}{T}=2 $
$ \Rightarrow $ $ T=\frac{10}{2}=5\,\min $