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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?

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superman1987

Answer:

(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.