Question

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

Answer 1

Answer:

Iron remains = 17.49 mg

Explanation:

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

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

Where, k is rate constant

So,  

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

$k=\frac {ln\ 2}{2.737}\ year^{-1}$

The rate constant, k = 0.2533 year⁻¹

Time = 2.41 years

$[A_0]$ = 32.2 mg

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

So,  

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

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

$[A_t]=17.49\ mg$

Iron remains = 17.49 mg