Question

3. Gold-198 decays with a half-life of 2.7 days according to the equation:198Au → 0−1β + 198Cd

Answer 1

a. k=0.256/day

b.sample of Au-198 remains after seven days : 1.67 g

Further explanation

1. A half-life of 2.7 days⇒t1/2=2.7 days

The half-life can be expressed in a decay constant( λ)

$\tt \displaystyle t_ {1/2} = {\dfrac {\ln (2)} {\lambda}}\\\\\lambda(or~k)=\dfrac{0.693}{2.7}=0.256/day$

2. We can use formula : (integrated rate law) :

$\tt ln[A]=-kt+ln[Ao]\\\\ln(\dfrac{Ao}{A})=kt$

Ao=10 g

t=7 days

k=0.256/day

$\tt ln(\dfrac{10}{A})=0.256\times 7\\\\ln(\dfrac{10}{A})=1.792\\\\e^{1.792}=\dfrac{10}{A}\\\\6=\dfrac{10}{A}\rightarrow A=1.67~g$