Question

The half-life of gold-198 is 2.7 days. After

Answer 1

Answer: 8.1 days

Explanation:

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

t = age of sample

a = let initial amount of the reactant = x

a - x = amount left after decay process= $\frac{x}{4}$ 

a) to find rate constant

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

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

$k=\frac{0.693}{2.7days}=0.257days^{-1}$

b) for completion of one fourth of reaction

$t=\frac{2.303}{k}\log\frac{x}{\frac{x}{4}}$

$t=\frac{2.303}{0.257}\log{4}$

$t=8.1days$

Thus after 8.1 days , one fourth of original amount will remain.