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:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
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).
<u><em>(a) 6 years:</em></u>
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]).
<em>Taking exponential for both sides:</em>
∴ 1.486 = ((1.1 x 10¹⁰ atoms)/[A]).
∴ [A] = (1.1 x 10¹⁰ atoms)/(1.486) = 7.4 x 10⁹ atoms.
<u><em>(b) 10 years</em></u>
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]).
<em>Taking exponential for both sides:</em>
∴ 1.935 = ((1.1 x 10¹⁰ atoms)/[A]).
∴ [A] = (1.1 x 10¹⁰ atoms)/(1.935) = 5.685 x 10⁹ atoms.
<u><em>(c) 200 years:</em></u>
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]).
<em>Taking exponential for both sides:</em>
∴ 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.
