Interpret the correct nuclear equation from following: 4 Be 7 → 3 Li 7 + -1 e 0 5 B 8 → 4 Be 8 + 1 e 0 94 Pu 239 + 2 He 4 → 96 Cm 242 + 1 H 1 43 Tc 97 + -1 e 0 → 42 Mo 97 4 Be 7 + -1 e 0 → 3 Li 7 . Concept Introduction: The decaying of an atom takes place by three type of decay such as alpha decay, beta decay, gama decay. In alpha decay, an alpha particle with mass 4m and electron 2e removed from nuclide while in beta decay two type of decay are possible beta (+) and beta (-). In gamma decay, nuclide will remain same to the original one.

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Answer 1

Question

Chapter 19, Problem 105CP

Interpretation Introduction

Interpretation:

Interpret the correct nuclear equation from following:

$4Be7  → 3Li7+-1e0$

$5B8    →  4Be8+1e0$

$94Pu239+2He4  → 96Cm242+1H1$

$43Tc97+-1e0  → 42Mo97$

$4Be7+-1e0  → 3Li7$ .

Concept Introduction:

The decaying of an atom takes place by three type of decay such as alpha decay, beta decay, gama decay.

In alpha decay, an alpha particle with mass 4m and electron 2e removed from nuclide while in beta decay two type of decay are possible beta (+) and beta (-). In gamma decay, nuclide will remain same to the original one.

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Answer 2

Question

Chapter 19, Problem 105CP

Interpretation Introduction

Interpretation:

Interpret the correct nuclear equation from following:

$4Be7  → 3Li7+-1e0$

$5B8    →  4Be8+1e0$

$94Pu239+2He4  → 96Cm242+1H1$

$43Tc97+-1e0  → 42Mo97$

$4Be7+-1e0  → 3Li7$ .

Concept Introduction:

The decaying of an atom takes place by three type of decay such as alpha decay, beta decay, gama decay.

In alpha decay, an alpha particle with mass 4m and electron 2e removed from nuclide while in beta decay two type of decay are possible beta (+) and beta (-). In gamma decay, nuclide will remain same to the original one.

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Answer 3

Question

Chapter 19, Problem 105CP

Interpretation Introduction

Interpretation:

Interpret the correct nuclear equation from following:

$4Be7  → 3Li7+-1e0$

$5B8    →  4Be8+1e0$

$94Pu239+2He4  → 96Cm242+1H1$

$43Tc97+-1e0  → 42Mo97$

$4Be7+-1e0  → 3Li7$ .

Concept Introduction:

The decaying of an atom takes place by three type of decay such as alpha decay, beta decay, gama decay.

In alpha decay, an alpha particle with mass 4m and electron 2e removed from nuclide while in beta decay two type of decay are possible beta (+) and beta (-). In gamma decay, nuclide will remain same to the original one.

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Answer 4

Question

Chapter 19, Problem 105CP

Interpretation Introduction

Interpretation:

Interpret the correct nuclear equation from following:

$4Be7  → 3Li7+-1e0$

$5B8    →  4Be8+1e0$

$94Pu239+2He4  → 96Cm242+1H1$

$43Tc97+-1e0  → 42Mo97$

$4Be7+-1e0  → 3Li7$ .

Concept Introduction:

The decaying of an atom takes place by three type of decay such as alpha decay, beta decay, gama decay.

In alpha decay, an alpha particle with mass 4m and electron 2e removed from nuclide while in beta decay two type of decay are possible beta (+) and beta (-). In gamma decay, nuclide will remain same to the original one.

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Answer 5

Chapter 19, Problem 105CP

Interpretation Introduction

Interpretation:

Interpret the correct nuclear equation from following:

$4Be7  → 3Li7+-1e0$

$5B8    →  4Be8+1e0$

$94Pu239+2He4  → 96Cm242+1H1$

$43Tc97+-1e0  → 42Mo97$

$4Be7+-1e0  → 3Li7$ .

Concept Introduction:

The decaying of an atom takes place by three type of decay such as alpha decay, beta decay, gama decay.

In alpha decay, an alpha particle with mass 4m and electron 2e removed from nuclide while in beta decay two type of decay are possible beta (+) and beta (-). In gamma decay, nuclide will remain same to the original one.

Answer 6

Chapter 19, Problem 105CP

Interpretation Introduction

Interpretation:

Interpret the correct nuclear equation from following:

$4Be7  → 3Li7+-1e0$

$5B8    →  4Be8+1e0$

$94Pu239+2He4  → 96Cm242+1H1$

$43Tc97+-1e0  → 42Mo97$

$4Be7+-1e0  → 3Li7$ .

Concept Introduction:

The decaying of an atom takes place by three type of decay such as alpha decay, beta decay, gama decay.

In alpha decay, an alpha particle with mass 4m and electron 2e removed from nuclide while in beta decay two type of decay are possible beta (+) and beta (-). In gamma decay, nuclide will remain same to the original one.

Answer 7

Chapter 19, Problem 105CP

Interpretation Introduction

Interpretation:

Interpret the correct nuclear equation from following:

$4Be7  → 3Li7+-1e0$

$5B8    →  4Be8+1e0$

$94Pu239+2He4  → 96Cm242+1H1$

$43Tc97+-1e0  → 42Mo97$

$4Be7+-1e0  → 3Li7$ .

Concept Introduction:

The decaying of an atom takes place by three type of decay such as alpha decay, beta decay, gama decay.

In alpha decay, an alpha particle with mass 4m and electron 2e removed from nuclide while in beta decay two type of decay are possible beta (+) and beta (-). In gamma decay, nuclide will remain same to the original one.

Answer 8

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Answer 10

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Answer 11

Ch. 19.1 - Prob. 1CT Ch. 19.1 - Prob. 19.1SC Ch. 19.1 - Prob. 19.2SC Ch. 19.3 - Exercise19.3Watches with numerals that glow in the...Ch. 19.8 - Prob. 1CT Ch. 19 - Prob. 1ALQ Ch. 19 - Prob. 2ALQ Ch. 19 - Prob. 3ALQ Ch. 19 - Prob. 4ALQ Ch. 19 - Prob. 5ALQ

Solution Summary: The author explains that the nuclear equation is balanced as the total number of atoms.

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