Q: A radioactive substance has 6.0 × 1018 active nuclei initially. What time is required for the active nuclei initially. What time is required the active nuclei of the same substance to become 1.0 × 1018 if its half – life is 40 s.
Sol: The number of active nuclei at any instant of time t,
$\large \frac{N_0}{N} = e^{\lambda t}$
$\large log_e\frac{N_0}{N} = \lambda t $
In this problem, the initial number of active nuclei, N0 = 6.0 × 1018 ; N = 1.0 × 1018 ,T= 40s,
$\large \lambda = \frac{0.693}{T}$
$\large \lambda = \frac{0.693}{40}$
= 1.733 × 10-2 s-1
$\large t = 2.303 log_{10}\frac{N_0}{N}/\lambda $
$\large t = 2.303 log_{10}\frac{N_0}{N}/\lambda $
$\large t = 2.303 log_{10}\frac{6 \times 10^{18}}{1 \times 10^{18}}/( 1.733 \times 10^{-2}) $
= (2.303 × 0.7782)/(1.733 × 10-2)
= 103.4 s.