Q132 / 180 · JIPMER 2015 · Physics

The half-life of ${ }^{198} \mathrm{Au}$ is 2.7 days. The average life is

  • A
    4 days
  • B
    3.4 days
  • Correct answer
    C
    3.9 days
  • D
    None of the above

Solution

The half-life and the decay constant are related as $\begin{aligned} t_{12} & =\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda} \\ \Rightarrow \quad \lambda & =\frac{0.693}{t_{1 / 2}}=\frac{0.693}{2.7 \mathrm{day}} \\ & =\frac{0.693}{2.7 \times 24 \times 3600 \mathrm{~s}} \\ & =2.9 \times 10^{-6} \mathrm{~s}^{-1}\end{aligned}$ Now average life $t_{a v}=\frac{1}{\lambda}=3.9$ days

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

Relevant Questions