Chapter 14, Problem 14.36PAE

Interpretation Introduction

Interpretation: Starting ${}^{\text{238}}\text{U}$ and ending with an isotope of Pb, write a decay scheme in which every nuclide emits only an $\text{α}$ particle .Compare your scheme to the decays series in Figure 14.5 and suggest a reason why ${\text{β}}^{-}$ emission must occur in the ${}^{\text{238}}\text{U}$ decay series.

Concept introduction:

Isotopes:

Isotopes are atoms of the same element which contains the same number of protons but different number of neutrons.

For example:

$\begin{array}{l}\text{Isotopes of Carbon}\\ {}_{\text{6}}^{\text{12}}{\text{C }}_{\text{6}}^{\text{13}}{\text{C }}_{\text{ 6}}^{\text{ 14}}\text{C}\end{array}$

Isotopes of atoms with unstable nuclei are called radioisotopes.

Radioactive decay: Unstable isotopes emit particles and release energy to become stable isotopes.

An element undergoes radioactive decay process it changes into a different isotope of the same element or into an entirely different element.

Alpha decay ( $\text{α}$): Nucleus emits an alpha particle containing two protons and two neutrons -helium nucleus. Transformation occurs as a nucleus has four less particles.

For example:

${}_{\text{64}}^{\text{149}}\text{Gd }\to {}_{\text{62}}^{\text{145}}{\text{Sm + }}_{\text{2}}^{\text{4}}\text{He}$

Beta Decay ( $\text{β}$):

A state where the nucleus of an atom has too many neutrons. The neutron transforms into a proton. An electron is released from the nucleus.

${}_{\text{6}}^{\text{14}}\text{C }\to {}_{\text{7}}^{\text{14}}{\text{N + }}_{\text{-1}}^{\text{0}}\text{e}$

Solution:

In the alpha emission unstable nuclides are formed. But in the beta emission stable nuclides are formed.