Natural decay process that the nuclide will undergo that has too large neutron-to-proton ratio has to be given.

Question

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Answer 1

Question

Chapter 21, Problem 21.1QE

Interpretation Introduction

Interpretation:

Natural decay process that the nuclide will undergo that has too large neutron-to-proton ratio has to be given.

Expert Solution & Answer

Explanation of Solution

If a nuclide has a large neutron-to-proton ratio means, then the number of neutrons is high than the number of protons present in the nucleus. In order to gain stability of the nucleus, this nuclide will undergo beta decay. In this natural process, a neutron is converted into a proton along with beta particle emission. Due to this the number of protons present in the nucleus will increase and neutrons will decrease. This can be represented in form of equation as,

n01→p11+β−10

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Answer 2

Question

Chapter 21, Problem 21.1QE

Interpretation Introduction

Interpretation:

Natural decay process that the nuclide will undergo that has too large neutron-to-proton ratio has to be given.

Expert Solution & Answer

Explanation of Solution

If a nuclide has a large neutron-to-proton ratio means, then the number of neutrons is high than the number of protons present in the nucleus. In order to gain stability of the nucleus, this nuclide will undergo beta decay. In this natural process, a neutron is converted into a proton along with beta particle emission. Due to this the number of protons present in the nucleus will increase and neutrons will decrease. This can be represented in form of equation as,

n01→p11+β−10

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Answer 3

Question

Chapter 21, Problem 21.1QE

Interpretation Introduction

Interpretation:

Natural decay process that the nuclide will undergo that has too large neutron-to-proton ratio has to be given.

Expert Solution & Answer

Explanation of Solution

If a nuclide has a large neutron-to-proton ratio means, then the number of neutrons is high than the number of protons present in the nucleus. In order to gain stability of the nucleus, this nuclide will undergo beta decay. In this natural process, a neutron is converted into a proton along with beta particle emission. Due to this the number of protons present in the nucleus will increase and neutrons will decrease. This can be represented in form of equation as,

n01→p11+β−10

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Answer 4

Question

Chapter 21, Problem 21.1QE

Interpretation Introduction

Interpretation:

Natural decay process that the nuclide will undergo that has too large neutron-to-proton ratio has to be given.

Expert Solution & Answer

Explanation of Solution

If a nuclide has a large neutron-to-proton ratio means, then the number of neutrons is high than the number of protons present in the nucleus. In order to gain stability of the nucleus, this nuclide will undergo beta decay. In this natural process, a neutron is converted into a proton along with beta particle emission. Due to this the number of protons present in the nucleus will increase and neutrons will decrease. This can be represented in form of equation as,

n01→p11+β−10

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 21, Problem 21.1QE

Interpretation Introduction

Interpretation:

Natural decay process that the nuclide will undergo that has too large neutron-to-proton ratio has to be given.

Expert Solution & Answer

Explanation of Solution

If a nuclide has a large neutron-to-proton ratio means, then the number of neutrons is high than the number of protons present in the nucleus. In order to gain stability of the nucleus, this nuclide will undergo beta decay. In this natural process, a neutron is converted into a proton along with beta particle emission. Due to this the number of protons present in the nucleus will increase and neutrons will decrease. This can be represented in form of equation as,

n01→p11+β−10

Answer 6

Chapter 21, Problem 21.1QE

Interpretation Introduction

Interpretation:

Natural decay process that the nuclide will undergo that has too large neutron-to-proton ratio has to be given.

Expert Solution & Answer

Explanation of Solution

If a nuclide has a large neutron-to-proton ratio means, then the number of neutrons is high than the number of protons present in the nucleus. In order to gain stability of the nucleus, this nuclide will undergo beta decay. In this natural process, a neutron is converted into a proton along with beta particle emission. Due to this the number of protons present in the nucleus will increase and neutrons will decrease. This can be represented in form of equation as,

n01→p11+β−10

Answer 7

Chapter 21, Problem 21.1QE

Interpretation Introduction

Interpretation:

Natural decay process that the nuclide will undergo that has too large neutron-to-proton ratio has to be given.

Answer 8

Expert Solution & Answer

Explanation of Solution

If a nuclide has a large neutron-to-proton ratio means, then the number of neutrons is high than the number of protons present in the nucleus. In order to gain stability of the nucleus, this nuclide will undergo beta decay. In this natural process, a neutron is converted into a proton along with beta particle emission. Due to this the number of protons present in the nucleus will increase and neutrons will decrease. This can be represented in form of equation as,

n01→p11+β−10