Chapter 19, Problem 55A

Interpretation Introduction

Interpretation :

The decay of uranium-238 tolead-206 should be explained.

Concept Introduction :

The decaying process in a radioactive element is occurring due to the type of processes such as alpha decay, beta decay and gamma decay.

Alpha decay involves the removal of alpha particle wit mass 4m and 2e electron from the nuclide. Bet decay is of the two types. The first one is beta-positive decay and the second one is bet negative decay. But in the gamma decay, the nuclide is same as the initial one.

Answer to Problem 55A

  ${}_{84}{\text{Po}}^{214}$ is the final nuclide which is the most stable nuclide which formed at the end of the decay process. It does not undergo any type of decay.

Explanation of Solution

The decay of uranium-238 to lead-26 is,

  ${}_{92}{\text{U}}^{238}{\stackrel{\alpha }{\to }}_{90}{\text{Th}}^{234}\stackrel{2\beta }{\to }{}_{92}{\text{U}}^{234}{\stackrel{5\alpha }{\to }}_{82}{\text{Pb}}^{214}{\stackrel{2\beta }{\leftarrow }}_{84}{\text{Po}}^{214}$

In this process, firstly uranium-238 undergoes the alpha decay, due to which the thorium-234 is produced by the reduction of 4 mass numbers and 2 atomic numbers which is unstable. Then this unstable thorium-234 radioactive element undergoes the beta decay, due to which the atomic number is increased by two factor and the product radioactive element is U-234. The third step, this unstable U-234 element isotope is undergoingthe 5 times alpha decay and the resulted element is Pb-214 and so, in the last step the stable

Nuclei ${}_{84}{\mathrm{Po}}^{214}$ is formed by undergoing 2 times beta decay. So, at last the ${}_{84}{\mathrm{Po}}^{214}$ nucleus is formed which is the stable nuclei.

Conclusion

Finally produced nuclide is the most stable nuclide.