# 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.

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.