If the amount of radioactive sodium-24, used for studies of electrolytes within the body, in a sample decreases from 0.8 to 0.2 mg in 30.0 h , what is the half-life of sodium-24?

15 h

Explanation:

Okay, the first thing that we all have to know before we can answer this question is that this Topic that is, Chemistry of Radioactivity is related to kinetics in a way that Radioactive disintegration follows the first order of Reaction which is under kinetics. So, we will be using the first order kinetics rate law to answer this question. Using the equation (1) below;

k =[ 2.303/ t ]×log ([N°}/ [Nr]) (1).

We are given from the question that N° = initial sample = 0.8 mg and Nr= sample remaining = 0.2 and the time taken = t= 30.0 h.

k= (2.303/ 30.0 h ) × log (0.8/0.2).

k=0.076768 h^-1 × log (4).

k= 0.076768 h^-1 × 0.6021.

k= 0.0462 h^-1.

Therefore, using the formula for Calculating half life below for first order kinetics we will be able to find out answer.

k = ln 2/ t(1/2). Where t(1/2) is the half life.

t(1/2) = ln 2/ k.

t(1/2) = ln 2 / 0.0462 h^-1.

t(1/2)= 0.6931/0.0462 h^-1.

t(1/2)=15 h

if you mean hydrogen and oxygen then water is the product