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Radioactive decay can be described by the following equation where is the original amount of the substance, is the amount of the substance remaining after time , and is a constant that is characteristic of the substance. For the radioactive isotope iron-55, is If the original amount of iron-55 in a sample is 32.2 mg, how much iron-55 remains after 2.41 years have passed? mg

Respuesta :

Answer:

Iron remains = 17.49 mg

Explanation:

Half life of iron -55 = 2.737 years (Source)

[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]

Where, k is rate constant

So,  

[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]

[tex]k=\frac {ln\ 2}{2.737}\ year^{-1}[/tex]

The rate constant, k = 0.2533 year⁻¹

Time = 2.41 years

[tex][A_0][/tex] = 32.2 mg

Using integrated rate law for first order kinetics as:

[tex][A_t]=[A_0]e^{-kt}[/tex]

Where,  

[tex][A_t][/tex] is the concentration at time t

[tex][A_0][/tex] is the initial concentration

So,  

[tex][A_t]=32.2\times e^{-0.2533\times 2.41}\ mg[/tex]

[tex][A_t]=32.2\times e^{-0.610453}\ mg[/tex]

[tex][A_t]=17.49\ mg[/tex]

Iron remains = 17.49 mg

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