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Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692

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Section
BuyFindarrow_forward

Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter 8, Problem 68E
Textbook Problem
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The figure below represents the decay of N-16 by alpha emission. Count the protons (red) and neutrons (gray), and draw in the missing particle. Write a nuclear equation to represent this decay.

Chapter 8, Problem 68E, The figure below represents the decay of N-16 by alpha emission. Count the protons (red) and

Interpretation Introduction

Interpretation:

The missing particle is to be drawn and a nuclear equation that represents the given decay is to be written.

Concept Introduction:

In alpha decay, the nucleus emits a particle with mass number 4 and atomic number 2, which is equivalent to a helium nucleus.

In alpha decay, the mass number of the element is reduced by 4 units and the atomic number is decreased by 2 units.

Explanation of Solution

Given Information:

In the given figure, the grey balls represent the number of neutrons and the red balls represent the number of protons. In the given diagram, the nucleus contains 7 protons and 9 neutrons. Since the mass number is the sum of the number of protons and neutrons, the mass number of the given nucleus is 16. Since the number of protons is equal to the atomic number, the atomic number is 7. Hence, the given nucleus is of nitrogen (716N).

In the equation, it is shown that the given particle emits an alpha particle. On the product side, the particle emitted contains two neutrons and two protons, which gives the atomic number 2 and mass number 4

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Chemistry In Focus
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