Q: A radioactive substance has 6.0 × 10^{18} 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 × 10^{18} 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, N_{0} = 6.0 × 10^{18} ; N = 1.0 × 10^{18} ,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.