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Chapter 18: Problem 4

How are the mass and the atomic number of a nucleus affected by the loss of a beta particle?

Short Answer

Expert verified
Mass number remains the same; atomic number increases by 1.

Step by step solution

01

Identify the Particles Involved

Understand that a beta particle is an electron (with a negative charge) that is emitted from the nucleus during the process called beta decay.
02

Understand Beta Decay

In beta decay, a neutron in the nucleus is converted into a proton and an electron. The electron (beta particle) is emitted from the nucleus.
03

Mass Number Analysis

The mass number (A) of the nucleus is the sum of protons and neutrons. Since a neutron is converted into a proton, there is no change in the total mass number. Thus, the mass number remains the same.
04

Atomic Number Analysis

The atomic number (Z), which is the number of protons in the nucleus, increases by 1 because a neutron is converted into a proton.

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Key Concepts

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

atomic number
The atomic number is one of the fundamental properties of an atom. It is the number of protons present in the nucleus of an atom. This number determines the chemical element. For instance, an atom with 6 protons is always a carbon atom. When an atom undergoes beta decay, the atomic number changes.

During beta decay, a neutron in the nucleus is converted into a proton. Since the atomic number is the count of protons, the increase of one proton causes the atomic number to increase by 1. So, if the original atom had an atomic number of 6 (Carbon), after beta decay, its atomic number would become 7.
mass number
The mass number of an atom is the total number of protons and neutrons in its nucleus. This is different from the atomic number which only counts protons. For example, a carbon atom with 6 protons and 6 neutrons has a mass number of 12.

In the process of beta decay, although a neutron is converted into a proton, the total count of protons and neutrons remains unchanged. Hence, the mass number does not change. This is crucial to understand as it shows why the mass of an atom remains the same even though its identity (atomic number) changes.
neutron-proton conversion
Neutron-proton conversion is the core process involved in beta decay. Inside the nucleus, a neutron (which is neutral) changes into a proton (which is positively charged). This transformation also produces an electron (beta particle) which is then emitted.

Let's break it down:
  • Neutron: Neutral charge.
  • Proton: Positive charge.
  • Electron: Negative charge.
This reaction also conserves the number of nucleons (neutrons + protons). So although the type of particles changes, their total count does not.
emission of beta particle
The emission of a beta particle is a significant event in beta decay. A beta particle is simply an electron that is emitted from the nucleus. This happens when a neutron turns into a proton and releases an electron.

This emitted electron is what we see as the beta particle. It's important to note that this electron was not originally present as an electron but was a part of the neutron. The emission carries away a small fraction of the energy, but this change is key to why the atomic number increases and helps maintain charge balance during the conversion process.

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Most popular questions from this chapter

Potassium-42 is used to locate brain tumors. Its half-life is \(12.5\) hours. Starting with \(15.4 \mathrm{mg}\), what fraction will remain after 100 hours? If it was necessary to have at least \(1 \mu \mathrm{g}\) for a particular procedure, could you hold the original sample for 200 hours before using it?

The nuclide \(\frac{223}{8}\) Fr emits three particles losing 8 mass units and 3 atomic number units. Propose a radioactive decay series and write the symbol for the resulting nuclide.

Write nuclear equations for the beta decay of (a) \({ }_{13}^{28} \mathrm{Al}\) (b) \({ }_{93}^{239} \mathrm{~Np}\)

Many home smoke detectors use americunium-241 as a radiation source. In the smoke detector, \({ }^{241}\) Am emits alpha particles. These alpha particles are detected by the detector unless smoke particles block their passage. Thus, as long as there are no smoke particles, the detector gets a continuous stream of alpha particles hitting it; smoke causes a disruption in this stream and activates the alarm. Write out the nuclear equation for the decay of \({ }^{241}\) Am by alpha emission.

Rubidium-87, a beta emitter, is the product of positron emission. Identify (a) the product of rubidium- 87 decay (b) the precursor of rubidium- 87
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