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

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

Step-by-step 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* e^(-0.0156* 267)`

`[A_t]=180* e^(-0.0156* 267)\ mg`

`[A_t]=2.79\ mg`

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