The barium isotope 133Ba has a half-life of 10.5 years. A sample begins with 1.1*1010 133Ba atoms. How many are left after (a) 6 years, (b) 10 years, and (c) 200 years?

(a) 7.4 x 10⁹ atoms.

(b)

(c)

Explanation:

It is known that the decay of a radioactive isotope isotope obeys first order kinetics. Half-life time is the time needed for the reactants to be in its half concentration. If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2). Also, it is clear that in first order decay the half-life time is independent of the initial concentration. The half-life of 133-Ba = 10.5 years.

For, first order reactions:

k = ln (2) / (t1/2) = 0.693 / (t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k = 0.693 / (t1/2) = 0.693 / (10.5 years) = 0.066 year⁻¹.

Also, we have the integral law of first order reaction:

kt = ln ([A₀]/[A]),

where, k is the rate constant of the reaction (k = 0.066 year⁻¹).

t is the time of the reaction.

[A₀] is the initial concentration of (133-Ba) ([A₀] = 1.1 x 10¹⁰ atoms).

[A] is the remaining concentration of (133-Ba) ([A] = ? g).

(a) 6 years:

t = 6.0 years.

∵ kt = ln ([A₀]/[A])

∴ (0.066 year⁻¹) (6.0 year) = ln ((1.1 x 10¹⁰ atoms) / [A])

∴ 0.396 = ln ((1.1 x 10¹⁰ atoms) / [A]).

Taking exponential for both sides:

∴ 1.486 = ((1.1 x 10¹⁰ atoms) / [A]).

∴ [A] = (1.1 x 10¹⁰ atoms) / (1.486) = 7.4 x 10⁹ atoms.

(b) 10 years

t = 10.0 years.:

∵ kt = ln ([A₀]/[A])

∴ (0.066 year⁻¹) (10.0 year) = ln ((1.1 x 10¹⁰ atoms) / [A])

∴ 0.66 = ln ((1.1 x 10¹⁰ atoms) / [A]).

Taking exponential for both sides:

∴ 1.935 = ((1.1 x 10¹⁰ atoms) / [A]).

∴ [A] = (1.1 x 10¹⁰ atoms) / (1.935) = 5.685 x 10⁹ atoms.

(c) 200 years:

t = 200.0 years.

∵ kt = ln ([A₀]/[A])

∴ (0.066 year⁻¹) (200.0 year) = ln ((1.1 x 10¹⁰ atoms) / [A])

∴ 13.2 = ln ((1.1 x 10¹⁰ atoms) / [A]).

Taking exponential for both sides:

∴ 5.4 x 10⁵ = ((1.1 x 10¹⁰ atoms) / [A]).

∴ [A] = (1.1 x 10¹⁰ atoms) / (5.4 x 10⁵) = 2.035 x 10⁴ atoms.