Chapter 22, Problem 22.10SP

Interpretation Introduction

Interpretation:

One isotope decays by $\text{β}$ emission and other isotope of tungsten decay by $\text{α}$ emission, has to be determined.

Concept Introduction:

Radioactive Nuclei:  The radioactive nuclei higher neutron/proton ratios tend to emit by $\text{β}$ particles, while nuclei with lower neutron/proton ratios tend to undergo nuclear decay by positron emission, electron capture, or $\text{α}$ emission.

Answer to Problem 22.10SP

${}^{185}\text{W}$ Undergoes $\text{β}$ emission; ${}^{160}\text{W}$ undergoes $\text{α}$ emission.

Explanation of Solution

$\begin{array}{l}{}^{160}\text{W (Atomic Number = 74)}{}^{185}\text{W (Atomic Number = 74)}\\ \text{ Neutron = (160-74) = 86}\text{Neutron = (185-74) = 111 }\\ \text{Proton = 74}\text{Proton = 74 }\\ \frac{\text{N}}{\text{P}}\text{= }\frac{86}{74}\text{ = 1}\text{.16}\frac{\text{N}}{\text{P}}\text{= }\frac{111}{74}\text{ = 1}\text{.5 }\end{array}$

Out of the given two isotopes of Tungsten, neutron/proton ratio of ${}^{160}\text{W}$ is less than the ratio value of ${}^{185}\text{W}$. Out of the two isotopes the neutron value of ${}^{185}\text{W}$ is greater and also its neutron/proton ratio; so, it undergoes $\text{β}$ emission.

The neutron value of ${}^{160}\text{W}$ less and also the ratio value; so, it undergoes $\text{α}$ emission.

Therefore, ${}^{185}\text{W}$ Undergoes $\text{β}$ emission; ${}^{160}\text{W}$ undergoes $\text{α}$ emission.