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In the formula A(t) = A0ekt, A(t) is the amount of radioactive material remaining from an initial amount A0 at a given time t and k is a negative constant determined by the nature of the material. A certain radioactive isotope decays at a rate of 0.1% annually. Determine the half-life of this isotope, to the nearest year.

Respuesta :

Answer:  693 years

Step-by-step explanation:

Expression for rate law for first order kinetics is given by:

[tex]A(t) =A_0e^{kt}[/tex]

k = rate constant

t = time taken for decomposition = 1

[tex]A_0[/tex] = Initial amount of the reactant

[tex]A_t[/tex] = amount of the reactant left =[tex]A_0-\frac{0.1}{100}\times A_0=0.999A_0[/tex]

[tex]0.999A_0=A_0e^{k\times 1}[/tex]

[tex]0.999=e^k[/tex]

[tex]k=-0.001year^{-1}[/tex]

for half life : [tex]t=t_\frac{1}{2}[/tex]

[tex]A_t=\frac{1}{2}A_o[/tex]

Putting in the values , we get

[tex]\frac{1}{2}A_0=A_0e^{-0.001\times t_\frac{1}{2}[/tex]

[tex]t_\frac{1}{2}=693 years[/tex]

Thus half life of this isotope, to the nearest year is 693.

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