Chapter 20.3, Problem 3PPB

Interpretation Introduction

Interpretation: From the given details of the wooden object, ${}^{14}C$ activity of old wooden object 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.

To determine: ${}^{14}C$ activity of old wooden object in a 2500 year old