EAMCET Medical EAMCET Medical Solved Paper-1996

  • question_answer
    The half-life of a radioactive isotope is 3 hours. The value of its disintegration constant is :

    A)  \[2.079\text{ hou}{{\text{r}}^{-1}}\]    

    B)  \[231\,\text{hou}{{\text{r}}^{-1}}\]

    C)  \[0.231\text{ hou}{{\text{r}}^{-1}}\]    

    D)  \[2.079\text{ hou}{{\text{r}}^{-1}}\]

    Correct Answer: C

    Solution :

                     Half life \[{{t}_{1/2}}=3\,hr.,\lambda =?\] \[\lambda =\frac{0.693}{{{t}_{t/2}}}=\frac{0.693}{3}=0.231\,\text{hou}{{\text{r}}^{-1}}\]


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