Q138 / 160 · TS EAMCET 2012 · Physics

The half-life of a radioactive element is $10 \mathrm{~h}$. The fraction of initial radioactivity of the element that will remain after $40 \mathrm{~h}$ is

  • A
    $\frac{1}{2}$
  • Correct answer
    B
    $\frac{1}{16}$
  • C
    $\frac{1}{8}$
  • D
    $\frac{1}{4}$

Solution

After 4 half-lifes $\frac{N_0}{2^4}=\frac{N_0}{16}$

Hence, the correct answer is option: 'B'.

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