If t3/4 and t1/4 are time required for completion of ¾ decay and ¼ decay, t3/4=t1/4×n.nthen is

Solution:

If t_3/4 and t_1/2 are time required for completion of 3/4 decay and 1/4 decay then t_3/4=t_1/2×n, then n is 2.

$For 3 / 4 decay, t_{3 / 4} = \frac{2.303}{k} \log \frac{100}{25} = \frac{1.39}{k} For 1 / 2 decay, t_{1 / 2} = \frac{2.303}{k} \log \frac{100}{50} = \frac{0.693}{k} n = \frac{t_{3 / 4}}{t_{1 / 2}} = \frac{1.39}{0.693} = 2$

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