The new nuclide formed is to be identified. Concept Introduction: A decaying nuclide emits an alpha particle by removing two protons and two neutrons. In this process, the mass numberand atomic number are conserved. The mass number and atomic number of the reactants are equal to the mass numbers and atomic numbers of the products.

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Chapter 15, Problem 113QP

Interpretation Introduction

Interpretation:

The new nuclide formed is to be identified.

Concept Introduction:

A decaying nuclide emits an alpha particle by removing two protons and two neutrons. In this process, the mass numberand atomic number are conserved. The mass number and atomic number of the reactants are equal to the mass numbers and atomic numbers of the products.

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