Chapter 20.3, Problem 20.3.2SR

Interpretation Introduction

Interpretation: From the age of the given wooden artifact, ${}^{14}C$ activity should be determined.

Concept Introduction:

• Unstable nuclei emit radiation spontaneously to become stable nuclei by losing energy. This process of emission of radiation by unstable nuclei is known as radioactive decay.
• These emitted radiations may be alpha radiations($\text{α}$), beta radiations($\text{β}$) or gamma radiations($\text{γ}$)
• These unstable nuclei are the nuclei with more than 83 protons and which do not lie within the belt of stability.
• Radioactive decay are in the first order kinetics. Rate of radioactive decay at a time t is,

$\begin{array}{l}Rateofdecayatatime\text{'}t\text{'}=kN\\ \text{k-}\text{First}\text{order}\text{rate}{\text{constant and its unit is t}}^{-1}\\ \text{N-}\text{number}\text{of}\text{radioactive}\text{nuclei}\text{present}\text{at}\text{time}\text{'t}\end{array}$

Suppose the number of radioactive nuclei at time zero is ${\text{N}}_{\text{0}}$ and at a time t is ${\text{N}}_{\text{t}}$,

$ln\frac{N_t}{N_0}=-kt$

• Half-life of radioactive decay is the time required for a radioactive sample to decay to one half of the atomic nucleus.

Half-life of the radiation,  $t_{1/2}=\frac{0.693}{k}$

Half-life and rate constant for radioactive isotopes vary greatly from nucleus to nucleus.

Answer to Problem 20.3.2SR

${}^{\text{14}}\text{C Activity of wooden artifact }\text{is 0}\text{.935}\text{disintegration}\text{per second}$

Explanation of Solution

Determine the value of k

Half-life of radioactive decay is the time required for a radioactive sample to decay to one half of the atomic nucleus. From the half-life of the reaction, the value of k can be easily determined.

$\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{\text{0}\text{.693}}{\text{k}}\\ \text{k}\text{=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}\end{array}$

Here the half lie of carbon-14 is 5715 years. So,

$\begin{array}{l}\text{k}\text{=}\frac{\text{0}\text{.693}}{\text{5715 years}}\\ \text{= 1}{\text{.21×10}}^{\text{-4}}{\text{yr}}^{\text{-1}}\end{array}$

Determine the ${}^{14}C$ activity of old wooden artifact in a $23000$ years old

Suppose the number of radioactive nuclei at time zero is ${\text{N}}_{\text{0}}$ and at a time t is ${\text{N}}_{\text{t}}$,

$ln\frac{N_t}{N_0}=-kt$

Here ${}^{14}C$ activity of the fresh cut wood is given. Since the radioactive sample is proportional to the number of radioactive nuclei, the above equation can be written with activity in place of concentration.

$\text{ln}\frac{{}^{\text{14}}\text{C}\text{activity}\text{in}\text{old}\text{wooden}\text{object}}{{}^{\text{14}}\text{C}\text{activity}\text{in}\text{fress}\text{cut}\text{wood}}\text{=}\text{-kt}$

$\begin{array}{l}\\ \text{From the given data's}\\ {}^{\text{14}}\text{C}\text{ activity}\text{in fresh cut wood}\text{= 15}\text{.2disintegration}\text{per second}\\ {}^{\text{14}}\text{C activity}\text{in}\text{the}\text{wooden}\text{artifact}\text{= ?}\\ \text{k}\text{= 1}\text{.21}{\text{×10}}^{\text{-4}}{\text{yr}}^{\text{-1}}\\ \text{Age}\text{of}\text{the}\text{wooden artifact,t}\text{=23,000}\text{yrs}\end{array}$

By adding these two values,

$\begin{array}{l}\text{ln}\frac{{}^{\text{14}}\text{C}\text{activity}\text{in}\text{old}\text{wooden}\text{object}}{\text{15}\text{.2disintegration}\text{per second}}\text{=}\text{-1}\text{.21}{\text{×10}}^{\text{-4}}{\text{yr}}^{\text{-1}}\text{×}\text{23000}\text{yr}\\ \text{=- 2}\text{.783}\\ \frac{{}^{\text{14}}\text{C}\text{activity}\text{in}\text{wooden}\text{artifact}}{\text{15}\text{.2}\text{dps}}\text{=}{\text{e}}^{\text{-2}\text{.783}}\\ {}^{\text{14}}\text{C}\text{activity}\text{in}\text{the}\text{wooden}\text{artifact}\text{=}{\text{e}}^{\text{-2}\text{.783}}\text{×15}\text{.2 dps}\\ \text{=}\text{0}\text{.940}\text{disintegration}\text{per second}\end{array}$

Conclusion

From the given details of the wooden object, ${}^{14}C$ activity is determined by using the equation of rate of decay.