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Q. 87.5% decomposition of a radioactive substance complete in 3 hours. What is the half-life of that substance?

MGIMS WardhaMGIMS Wardha 2003

Solution:

Given, $ {{N}_{0}}=100,\text{ }N=100-87.5=12.5, $ $ t=3\,hours,\,{{t}_{1/2}}=? $ We know that $ \frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{\frac{t}{{{t}_{1/2}}}}} $ $ \frac{12.5}{100}={{\left( \frac{1}{2} \right)}^{3/{{t}_{1/2}}}} $ or $ {{\left( \frac{1}{2} \right)}^{3}}={{\left( \frac{1}{2} \right)}^{3/{{t}_{1/2}}}} $ Hence, $ \frac{3}{{{t}_{1/2}}}=3 $ or $ {{t}_{1/2}}=\frac{3}{3}=1\text{ }hour $