If this decay has a half-life of 2.60 years, what mass of 72.5 g sodium-22 will remain after 15.6 years

1.1328 g left after 15.6 years

Explanation:

First we need to find, that how many times it decays: 15.6 ÷ 2.60 = 6 times

Then we know the formula: mass of original ÷ 2^n = remaining mass.

                                           :     72.5   ÷   2^6 = 1.1328 g left.

remaining mass = 1.13 grams (3-sig figs)

Explanation:

All radioactive decay follows 1st order kinetics and is defined by the expression ...

A = A₀e^⁻kt

A = final mass activity

A₀ = initial mass activity

k = rate constant = 0.693/t(half-life) = 0.693/2.60 yrs = 0.2665 yrs⁻¹

t = time of decay = 15.6 yrs

A = (72.5 g)(e^[-(0.2665 yrs⁻¹)(15.6yrs)] = 72.5 grams x 0.0156

= 1.13381331 grams (calc. ans.) ≅ 1.13 grams (3 sig figs)

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