Question

Iron-59 is a beta emitter with a half-life of 44.5 days. If a sample initially contains 180 mg of iron-59, how much iron-59 is left in the sample after 267 days?

Answer 1

Answer:

2.79 mg iron-59 is left in the sample after 267 days.

Explanation:

Given that:

Half life = 44.5 days

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

Where, k is rate constant

So,  

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

$k=\frac {ln\ 2}{44.5}\ days^{-1}$

The rate constant, k = 0.0156 days⁻¹

Using integrated rate law for first order kinetics as:

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

Where,  

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration  = 180 mg

Time = 267 days

So,  

$[A_t]=180\ mg\times e^{-0.0156\times 267}$

$[A_t]=180\times e^{-0.0156\times 267}\ mg$

$[A_t]=2.79\ mg$

2.79 mg iron-59 is left in the sample after 267 days.