/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 160 Loss of a \(\beta\)-particle is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Loss of a \(\beta\)-particle is equivalent to a. increase of one proton only b. decrease of one neutron only c. both a and \(\mathbf{b}\) d. none of these

Short Answer

Expert verified
c. both a and b

Step by step solution

01

Understanding Beta Decay

In beta decay, a neutron in an unstable nucleus is transformed into a proton, an electron (beta particle), and an antineutrino. The electron is emitted from the nucleus as a beta particle.
02

Effect on Atomic Number and Mass

When a beta particle is emitted, the atomic number of the element increases by 1 because a neutron has been converted into a proton. However, the mass number remains unchanged as the sum of protons and neutrons is the same.
03

Analyzing the Options

Given the process of beta decay:- a. The atomic number increases by one due to the addition of a proton.- b. A neutron is converted into a proton, effectively decreasing the neutron count by one.- c. Both a and b occur during beta decay.Therefore, option c ('both a and \(b\)') is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Neutron to Proton Conversion
During beta decay, a remarkable transformation occurs within an unstable atomic nucleus. Inside the nucleus, a neutron changes into a proton. This process is crucial because it alters the composition of the nucleus, but surprisingly, it doesn't change the mass number of the atom.

Now, you might wonder how a neutron can simply become a proton. Here's what happens: each neutron is made up of smaller particles called quarks. During beta decay, one of the quarks inside the neutron changes type, turning the neutron into a proton. This change emits an electron (known as a beta particle) and an antineutrino from the nucleus.

It's essential to understand that:
  • The proton count in the nucleus increases by one.
  • The neutron count decreases by one.
  • The overall mass of the nucleus remains the same because protons and neutrons have nearly identical mass.
This delicate transformation keeps the atomic mass constant while increasing the atomic number, resulting in a new element.
Atomic Number Increase
In the world of atoms, the atomic number is all about the number of protons in the nucleus. It's like the ID card for an element, determining its chemical properties and place on the periodic table.

When beta decay occurs, the atomic number goes up by one. Why? Because one neutron in the nucleus transforms into a proton, adding to the nuclear count of protons by one.

This increase in protons means that the atom becomes a different element, now further down the periodic table by one spot:
  • If we started with carbon-14, as a typical example, it becomes nitrogen-14 after the beta decay process.
  • The atomic number changes, here from 6 (for carbon) to 7 (for nitrogen).
This simple increase in atomic number changes the atomic identity, transforming it into a completely new element.
Mass Number Unchanged
One of the unique aspects of beta decay is that despite the changes going on inside the nucleus, the mass number remains unchanged. The mass number is the total count of protons and neutrons in an atom's nucleus.

During beta decay:
  • A neutron converts to a proton.
  • Because protons and neutrons have roughly the same mass, the total mass number, which is the sum of both, does not change.
  • The total number of nucleons (protons + neutrons) stays the same, although the actual composition shifts.
So, although the atom's identity changes due to a new proton emerging, the weight it carries stays constant. This property helps in maintaining balance within nuclear processes while allowing the atom to progress into new elements through decay events.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(A): The reactions taking place in the sun are nuclear fusion reactions. (R): The main reason for nuclear fusion reactions in the sun is that \(\mathrm{H}_{2}\) is present in the sun's atmosphere so that hydrogen nuclei can fuse to form helium.

Which of the following reaction represents \({ }_{22} \mathrm{Ti}^{44}\) decay by electron capture? a. \({ }_{22} \mathrm{~T}_{1}^{-44}+{ }_{-1} \mathrm{e}^{0} \rightarrow{ }_{21} \mathrm{Sc}^{44}\) b. \(\quad, T_{1}^{i 4}+, \mathrm{e}^{0} \rightarrow\) c. \(_{22} \mathrm{Ti}^{44}+{ }_{0} \mathrm{e}^{1} \rightarrow{ }_{22} \mathrm{Ti}^{45}\) d. \({ }_{22} \mathrm{Ti}^{44}+{ }_{0} \mathrm{e}^{-1} \rightarrow{ }_{22} \mathrm{Ti}^{43}\)

For the nuclei with number \(>100\) a. The nucleus essentially breaks up into two nuclides of equal mass releasing energy b. The two nuclei fuse to form a bigger nuclide releasing energy c. Binding energy of the nucleus increases on an average as A increases. d. Binding energy of the nucleus decreases on an average as A increases.

The number of \(\alpha\) and \(\beta\)-particle emitted in the nuclear reaction \({ }_{90}^{28} \mathrm{Th} \rightarrow{ }_{83}^{212} \mathrm{Bi}\) are a. \(4 a\) and \(1 \beta\) b. \(3 \alpha\) and \(7 \beta\) c. \(8 \alpha\) and \(1 \beta\) d. \(4 \alpha\) and \(7 \beta\)

Which of the following pairs are isodiapheric pairs? a. \({ }_{29} \mathrm{Cu}^{65}\) and \({ }_{24} \mathrm{Cr}^{55}\) b. \({ }_{29} \mathrm{Cu}^{65}\) and \({ }_{24} \mathrm{Cr}^{52}\) c. \({ }_{92} \mathrm{U}^{235}\) and \(_{90} \mathrm{Th}^{231}\) d. \({ }_{92} \mathrm{U}^{238}\) and \({ }_{90} \mathrm{Th}^{231}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.