Chapter U1, Problem SIII3RQ

Interpretation Introduction

Interpretation:

A nuclear equation for the beta decay of cadmium-113 should be determined.

Concept introduction: When an atom on undergoing radioactive decay emits a fast-moving electron from the nucleus of an atom then such type of radioactive decays is said to be beta emission.

Answer to Problem SIII3RQ

${}_{48}^{113}\text{Cd }\to {}_{49}^{113}{\text{In}}^{}{\text{+ }}_{-1}^0\text{β}$

Explanation of Solution

The general reaction for beta decay is:

${}_Z^A\text{X }\to {}_{Z+1}^A{\text{X}}^{}{\text{+ }}_{-1}^0\text{β}$

Where A is mass number, Z is atomic number (number of protons), and X is the symbol of element.

The given atom is cadmium-113.

Since, the atomic number of cadmium is 48 so, the formula of the element will be ${}_{48}^{113}\text{Cd}$. The beta emission will result in increase in number of protons by 1 so, the number of protons of the element formed from beta emission of cadmium-113will be 49 and the mass number will remain same that is 113. The element with atomic number 49 is Indium, In. Thus, the nuclear reaction of beta emitter of cadmium-113is:

${}_{48}^{113}\text{Cd }\to {}_{49}^{113}{\text{In}}^{}{\text{+ }}_{-1}^0\text{β}$