Question

A sample of ${U^{238}}$ (half-life $ = 4.5 \times {10^9}$ years) ore is found to contain $23.8g$ of ${U^{238}}$ and $20.6g$ of $P{b^{206}}$. The age of the ore is:
$
  A)4.5 \times {10^9}years \\
  B)3.75 \times {10^9}years \\
  C)6.25 \times {10^9}years \\
 $
$D)$None of these

Answer 1

Hint: The half-life of a species in a chemical reaction is the time it takes for its concentration to fall to half of its original value, or, equivalently, the time it takes for a radioactive material's number of disintegrations per second to decrease by one-half.

Complete step-by-step answer:
The mechanism through which an unstable atomic nucleus releases energy by radiation is known as radioactive decay (also known as nuclear decay, radioactivity, radioactive disintegration, or nuclear disintegration). The term "radioactive" refers to a substance that contains unstable nuclei.

The equation of decay is given by:
${}_{92}^{238}U \to {}_{82}^{206}Pb + 8{}_2^4He + 6{}_{ - 1}^0e$
We know the total number of gram atoms of $Pb$ present $ = \dfrac{{20.6}}{{206}} = 0.1g.atom$ . Total number of gram atoms of $Pb$ present corresponds to the total number of gram atoms of $U$ decayed.

We also know the total number of gram atoms of $U$ present $ = \dfrac{{23.8}}{{238}} = 0.1g.atom$
therefore, $N = 0.1g.atom$
${N_0} = $total number of g atoms of $U$ present $ + $ total number of gram atoms of $U$ decayed
${N_0} = 0.1 + 0.1 = 0.2g.atom$
The integrated rate law expression for the first-order decay process is :
$
  t = \dfrac{{2.303}}{\lambda }{\log _{10}}\dfrac{{{N_0}}}{N} = \dfrac{{2.303 \times half.life.period}}{{0.693}}{\log _{10}}\dfrac{{{N_0}}}{N} \\
  t = \dfrac{{2.303 \times 4.5 \times {{10}^9}}}{{0.693}}{\log _{10}}\dfrac{{0.2}}{{0.1}} \\
  t = 4.5 \times {10^9} \\
 $
Hence, the age of the ore is $t = 4.5 \times {10^9}$ years.

Additional Information: The five types of radioactivity are alpha decay, beta decay, gamma emission, positron emission, and electron capture. Rays are often used in nuclear reactions, and certain nuclei decay due to electron capture. Each one of these decay modes results in the creation of a new nucleus with a more stable neutron and proton ratio.

Hence, option $A$ is the correct answer.

Note: Radioactivity is an integral part of the internal dynamics of the earth as well as a useful method for geologists. It is the planet's primary source of deep heat, as well as the cause of its internal history and current dynamics.