Q31.5-54PE - College Physics (Urone)

Question

A tree falls in a forest. How many years must pass before the 14C activity in 1.00 g of the tree’s carbon drops to 1.00 decay per hour?

Step-by-Step Solution

Answer

The tree’s carbon drops to 1.00 decay per hour is \(5.62 \times {10^4}\,{\rm{y}}\).

Definition of tree’s carbon drop

The growth features of the tree species, the growing environment where the tree is planted, and the wood density of the tree all affect the rate of carbon sequestration. It is strongest when a tree is young, between 20 and 50 years old. The fact that far less research has been conducted on tropical tree species than on temperate tree species further exacerbates the problem.

Calculation

\(^{12}{\rm{C}}\)Is the mass of one mole of \(M  = 12\,\;{\rm{g}}\)atoms.

As a result, the number of atoms in an\(m  = 1.00\;\,{\rm{g}}\) atom is.

\(\begin{align}{\underline{\phantom{xx}}}{N_1} & = \frac{m}{M}{N_A}\\ & = \frac{{(1.00\,{\rm{g}})}}{{(12.0\,{\rm{g}})}}\left( {6.02 \times {{10}^{23}}\,{\rm{atoms }}} \right)\\ & = 5.02 \times {10^{22}}\,{\rm{atoms }}\end{align}\)

The natural occurrence of the \(^{14}{\rm{C}}\) atom is now\(1.3 \times {10^{ - 12}}\). As a result, the sample's initial number of \(^{14}{\rm{C}}\) atoms is:

\(\begin{align}{\underline{\phantom{xx}}}{N_i} & = \left( {1.3 \times {{10}^{ - 12}}} \right)\left( {5.02 \times {{10}^{22}}\,{\rm{atoms }}} \right)\\ & = 6.52 \times {10^{10}}\,{\rm{atoms }}\end{align}\)

The sample's current activity is decaying at \(R  = 1.0\)\(/{\rm{hr}}  = \frac{{1.0}}{{3600}}\,{\rm{decay/s}}\)

The total number of atoms in the sample is

 

\(\begin{align}{\underline{\phantom{xx}}}{N_f} & = \frac{{R{t_{1/2}}}}{{0.693}}\\ & = \frac{{\left( {\frac{1}{{3600}}\,{\rm{decay/s}}} \right)\left( {1.81 \times {{10}^{11}}\,{\rm{s}}} \right)}}{{0.693}}\\ & = 7.26 \times {10^7}\,{\rm{atoms }}\end{align}\)

Therefore, the amount of time that has passed is equal to

\(\begin{align}{\underline{\phantom{xx}}}t & = \frac{{\ln \left( {\frac{{{N_i}}}{{{N_f}}}} \right)}}{{0.693}}{t_{1/2}}\\ & = \frac{{\ln \left( {\frac{{6.52 \times {{10}^{10}}\,{\rm{atoms }}}}{{7.26 \times {{10}^7}\,{\rm{atoms }}}}} \right)}}{{0.693}}(5730\,{\rm{y}})\\ & = 5.62 \times {10^4}\,{\rm{y}}\end{align}\)

 

 

Hence, the tree’s carbon drops to 1.00 decay per hour is\(5.62 \times {10^4}\,{\rm{y}}\) .