Question

Half life (t,) for a radioactive decay is 6930 sec. The time
required to fall the rate of decay by
of 1/100 its initial value​

Answer 1

Answer: 46060 seconds

Explanation:

Half-life of radioactive isotope= 6930 sec

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_\frac{1}{2}}$

$k=\frac{0.693}{6930}=10^{-4}\text{s}^{-1}$

Now we have to calculate the age of the sample:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = [tex]10^{-4}\text{s}^{-1}

t = age of sample  = ?

a = let initial amount of the reactant  = a

a - x = amount left after decay process  = $\frac{1}{100}\times a$

Now put all the given values in above equation, we get

$t==\frac{2.303}{10^{-4}}\log\frac{a}{\frac{1}{100}\times a}$

$t=46060sec$

Thus the time  required to fall the rate of decay by  of 1/100 its initial value​ is 46060 seconds.