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The half life of Pb-210 is 22 years. A decayed animal shows 25% of the original Pb-210 remains; how long has the animal been deceased to the nearest tenth of a year?

Respuesta :

wolf1728
This is an easy half-life calculation.  (Most of these are NOT easy).
If 25% remains of an original amount of 100%, then the time elapsed is precisely 44 years.  (2 half-lives)
In 22 years (one half-life), it would be 50%
In 44 years (two half-lives), it would be 25%
In 66 years (three half-lives), it would be 12.5%
In 88 years (four half-lives), it would be 6.25%

Answer:

44 years.

Step-by-step explanation:

Let original amount be 100.

We have been given that the half life of Pb-210 is 22 years. A decayed animal shows 25% of the original Pb-210 remains.

We will use half-life formula to solve our given problem.

[tex]A=a\cdot (0.5)^{\frac{t}{h}}[/tex], where,

A = Amount after t units of time,

a = Initial amount,

t = Time,

h = Half-life.

[tex]25=100\cdot (0.5)^{\frac{t}{22}}[/tex]

[tex]\frac{25}{100}=\frac{100\cdot (0.5)^{\frac{t}{22}}}{100}[/tex]

[tex]0.25=(0.5)^{\frac{t}{22}}[/tex]

Take natural log of both sides:

[tex]\text{ln}(0.25)=\text{ln}((0.5)^{\frac{t}{22}})[/tex]

Using natural log property [tex]\text{ln}(a^b)=b\cdot \text{ln}(a)[/tex], we will get:

[tex]\text{ln}(0.25)=\frac{t}{22}*\text{ln}(0.5)[/tex]

[tex]\frac{\text{ln}(0.25)}{\text{ln}(0.5)}=\frac{t*\text{ln}(0.5)}{22*\text{ln}(0.5)}[/tex]

[tex]\frac{-1.38629436}{-0.69314718}=\frac{t}{22}[/tex]

[tex]2=\frac{t}{22}[/tex]

[tex]2*22=\frac{t}{22}*22[/tex]

[tex]44=t[/tex]

Therefore, the animal has been deceased to 44 years.

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