Question

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?

Answer 1

`t=23.07\ years`

Step-by-step explanation:

Half life of lead - 210 = 22.3 years

`t_(1/2)=\frac {ln\ 2}{k}`

Where, k is rate constant

So,

`k=(\ln2)/(t_(1/2))`

`k=(\ln2)/(22.3)\ year^(-1)`

The rate constant, k = 0.0311 year⁻¹

Using integrated rate law for first order kinetics as:

`[A_t]=[A_0]e^(-kt)`

Where,

`[A_t]` is the concentration at time t

`[A_0]` is the initial concentration

Initial concentration
`[A_0]` = 33.2 mg

Final concentration
`[A_t]` = 16.2 mg

Time = ?

Applying in the above equation, we get that:-

`16.2=33.2e^(-0.0311* t)`

`332e^(-0.0311t)=162`

`e^(-0.0311t)=(81)/(166)`

`-0.0311t=\ln \left((81)/(166)\right)`

`t=23.07\ years`