Question

The half-life of Erbium-165 is 10.4 hours. After 24 hours a sample has been reduced to a mass of 2 mg. What was the initial mass of the sample, and how much will remain after 3 days?

Answer 1

Answer:

$[A_0]=10\ mg$

$[A_t]=0.08245\ mg$

Explanation:

Given that:

Half life = 10.4 hours

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,  

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{10.4}\ hour^{-1}$

The rate constant, k = 0.06664 hour⁻¹

Time = 24 hours

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t  = 2 mg

$[A_0]$ is the initial concentration = ?

So,  

$2\ mg=[A_0]\times e^{-0.06664\times 24}$

$[A_0]=10\ mg$

Now, time = 3 days = 3*24 hours = 72 hours ( 1 day = 24 hours)

$[A_0]=10\ mg$

Thus,

$[A_t]=10\times e^{-0.06664\times 72}\ mg=0.08245\ mg$