The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?

0.0468 g.

Explanation:

The decay of radioactive elements obeys first-order kinetics. For a first-order reaction: k = ln2 / (t1/2) = 0.693 / (t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2 / (t1/2) = 0.693 / (1620 years) = 4.28 x 10⁻⁴ year⁻¹.

For first-order reaction: kt = lna / (a-x).

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna / (a-x)

(4.28 x 10⁻⁴ year⁻¹) (12960 year) = ln (12) / (a-x).

5.54688 = ln (12) / (a-x).

Taking e for the both sides:

256.34 = (12) / (a-x).

∴ (a-x) = 12/256.34 = 0.0468 g.