Question

The half life of Pb-210 is 22 years. A decayed animal shows 48% of the original Pb-210 remains; how long has the animal been deceased to the nearest tenth of a year?A .0.01 yearsB. 23.3 yearsC. 0.04 yearsD. 23.29 years

Answer 1

Step 1:

In this question, we are given the following:

The half-life of Pb-210 is 22 years.

A decayed animal shows 48% of the original Pb-210 remains;

how long has the animal been deceased to the nearest tenth of a year?

Step 2:

The details of the solution are as follows:

`\begin{gathered} From\text{ the question, we can see that:} \\ 0.\text{ 5 = e}^(22k) \\ Making\text{ k the subject of the formulae, we have that:} \\ k\text{ = \lparen}\frac{ln\text{ 0. 5}}{22})\text{ --- equation 1} \end{gathered}`
`\begin{gathered} \text{0. 48 = e}^{(\frac{I\text{n 0. 5}}{22})t} \\ Taki\text{ng In of both sides, we have that:} \\ ln\text{ \lparen0. 48 \rparen= \lparen}\frac{ln\text{ 0. 5}}{22}\text{ \rparen t} \end{gathered}`
`Multiply\text{ both sides by \lparen}\frac{22}{ln\text{ 0. 5 }})\text{ , we have that:}`
`\begin{gathered} \text{t = \lparen ln 0.48 \rparen }*\text{ \lparen}\frac{22}{ln\text{ 0. 5 }}) \\ t=\text{ 23.29566} \\ t\approx\text{ 23.29 years \lparen OPTION D \rparen} \end{gathered}`



CONCLUSION:

The final answer is:

`\text{t }\approx\text{ 23. 29 years \lparen OPTION D \rparen}`