Radioactive decay can be described by the following equation:
λ(s⁻¹) = ln (2) / T₁₂
, where s is the original amount of the substance, λ(s⁻¹) is the amount of the substance remaining after time T₁₂ and is a constant that is characteristic of the substance.
If the original amount of lead-210 in a sample is 33.2 mg, how much time is needed for the amount of lead-210 that remains to fall to 16.2 mg?

Respuesta :

Answer:

[tex]t=23.07\ years[/tex]

Explanation:

Half life of lead - 210 = 22.3 years

[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]

Where, k is rate constant

So,  

[tex]k=\frac{\ln2}{t_{1/2}}[/tex]

[tex]k=\frac{\ln2}{22.3}\ year^{-1}[/tex]

The rate constant, k = 0.0311 year⁻¹

Using integrated rate law for first order kinetics as:

[tex][A_t]=[A_0]e^{-kt}[/tex]

Where,  

[tex][A_t][/tex] is the concentration at time t

[tex][A_0][/tex] is the initial concentration

Initial concentration [tex][A_0][/tex] = 33.2 mg

Final concentration [tex][A_t][/tex] = 16.2 mg

Time = ?

Applying in the above equation, we get that:-

[tex]16.2=33.2e^{-0.0311\times t}[/tex]

[tex]332e^{-0.0311t}=162[/tex]

[tex]e^{-0.0311t}=\frac{81}{166}[/tex]

[tex]-0.0311t=\ln \left(\frac{81}{166}\right)[/tex]

[tex]t=23.07\ years[/tex]