Question

A radioactive element x converts into another stable element y. Half-life of x is 2 h, initially only x is present. After time r, the ratio of atoms of x and y is found to be 1 : 4, then t in hour is

• A. 2
• B. 4
• C. between 4 and 6
• D. 6

Answer 1

Answer: (C)

Let $ {{N}_{0}} $ be the number of atoms of $ x $ at time $ t=0 $ . Then at $ t=4h, $ ie, (two half-lives) $ {{N}_{x}}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{2}} $ $ {{N}_{x}}=\frac{{{N}_{0}}}{4} $ $ \therefore $ $ {{N}_{y}}={{N}_{0}}-\frac{{{N}_{0}}}{4}=\frac{3{{N}_{0}}}{4} $ $ \therefore $ $ \frac{{{N}_{x}}}{{{N}_{y}}}=\frac{1}{3} $ Now, at $ t=6\,h, $ ie, (three half-lives) $ {{N}_{x}}={{N}_{0}}{{\left( \frac{1}{2} \right)}^{3}}=\frac{{{N}_{0}}}{8} $ and $ {{N}_{x}}={{N}_{0}}-{{N}_{x}} $ $ ={{N}_{0}}-\frac{{{N}_{0}}}{8} $ $ =\frac{7{{N}_{0}}}{8} $ or $ \frac{{{N}_{x}}}{{{N}_{y}}}=\frac{1}{7} $ The given ratio lies between $ \frac{1}{3} $ and $ \frac{1}{7} $ Therefore, t lies between 4 h and 6 h.