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

Short Answer

Expert verified
The nuclear equations for the beta decay are: (a) \[ { }_{13}^{28} \mathrm{Al} \rightarrow { }_{14}^{28} \mathrm{Si} + \beta^- + \bar{u}_e \] (b) \[ { }_{93}^{239} \mathrm{Np} \rightarrow { }_{94}^{239} \mathrm{Pu} + \beta^- + \bar{u}_e \]

Step by step solution

01

Identify the initial element and the type of decay

In beta decay, a neutron is converted into a proton, and an electron (beta particle) is emitted. The atomic number increases by 1 while the mass number remains unchanged.
02

Write the nuclear equation for \({ }_{13}^{28} \mathrm{Al}\)

The initial element is aluminum-28, symbolized as \({ }_{13}^{28} \mathrm{Al}\). After beta decay, the new element will have an atomic number of 14 (silicon), while the mass number remains 28. The emitted beta particle is symbolized as \( \beta^- \). The equation is: \[ { }_{13}^{28} \mathrm{Al} \rightarrow { }_{14}^{28} \mathrm{Si} + \beta^- + \bar{u}_e \]
03

Write the nuclear equation for \({ }_{93}^{239} \mathrm{Np}\)

The initial element is neptunium-239, symbolized as \({ }_{93}^{239} \mathrm{Np}\). After beta decay, the new element will have an atomic number of 94 (plutonium), while the mass number remains 239. The equation is: \[ { }_{93}^{239} \mathrm{Np} \rightarrow { }_{94}^{239} \mathrm{Pu} + \beta^- + \bar{u}_e \]

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Key Concepts

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

beta decay
Beta decay is a type of radioactive decay where a neutron inside an atomic nucleus transforms into a proton and an electron. This means the element changes because the atomic number increases by 1, but the mass number stays the same. The emitted electron is called a beta particle, symbolized as \(\beta^-\).
This process also releases a very tiny particle with hardly any mass, called an antineutrino, symbolized as \(\bar{u}_e\). Beta decay is important because it shows how some elements can change into different elements naturally over time.
nuclear equations
Nuclear equations help us understand and describe beta decay in a structured way.
They follow certain rules to make sure everything balances out.
One key rule is that the sum of atomic numbers and the sum of mass numbers must be the same on both sides of the equation.
Let's look at the beta decay of aluminum-28. Starting with \({ }_{13}^{28} \text{Al}\), the equation would be: \[ { }_{13}^{28} \text{Al} \rightarrow { }_{14}^{28} \text{Si} + \beta^- + \bar{u}_e \]
Here, aluminum becomes silicon because the atomic number goes up from 13 to 14, but the mass number stays the same at 28.
Now, let's see the beta decay for neptunium-239: \[ { }_{93}^{239} \text{Np} \rightarrow { }_{94}^{239} \text{Pu} + \beta^- + \bar{u}_e \]
Here, neptunium turns into plutonium because its atomic number increases from 93 to 94, while the mass number remains at 239.
By writing nuclear equations like these, we can clearly show what happens during the process of beta decay.
atomic structure
To understand beta decay, we need to know about atomic structure.
An atom consists of three main particles: protons, neutrons, and electrons.
  • Protons have a positive charge.
  • Neutrons have no charge.
  • Electrons have a negative charge.

The protons and neutrons are in the nucleus at the center of the atom, while the electrons orbit around the nucleus.
In beta decay, a neutron in the nucleus changes into a proton and an electron (beta particle). This changes the element because the number of protons defines what element the atom is.
radioactivity
Radioactivity is the process by which unstable atomic nuclei lose energy by emitting radiation.
This can happen in several ways, including beta decay. When an atom undergoes beta decay, it releases a beta particle (electron) and an antineutrino.
  • These emissions are forms of radiation.
  • Radioactivity helps us understand how elements can transform naturally over time.

Studying radioactivity is important for several reasons.
  • It's crucial in nuclear medicine for treating and diagnosing diseases.
  • In geology, it helps date ancient rocks.
  • It's critical in nuclear power generation.

Knowing about radioactivity gives us a deeper understanding of natural processes and helps us harness them for practical uses.

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Most popular questions from this chapter

Consider the fission reaction $$ { }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{35}^{94} \mathrm{Sr}+{ }_{54}^{139} \mathrm{Xe}+3{ }_{0}^{1} \mathrm{n}+\text { energy } $$ Calculate the following using these mass data \((1.0 \mathrm{~g}\) is equivalent to \(9.0 \times 10^{13} \mathrm{~J}\) ): $$ \begin{array}{rlr} \mathrm{U}-235=235.0439 \mathrm{amu} & \text { Sr-94 }=93.9154 \mathrm{amu} \\\ \mathrm{Xe}-139=138.9179 \mathrm{amu} & \mathrm{n}=1.0087 \mathrm{amu} \end{array} $$ (a) the energy released in joules for a single event (one uranium atom splitting) (b) the energy released in joules per mole of uranium splitting (c) the percentage of mass lost in the reaction

The curie is equal to \(3.7 \times 10^{10}\) disintegration/sec, and the becquerel is equivalent to just 1 disintegration/sec. Suppose a hospital has a 150 -g radioactive source with an activity of \(1.24 \mathrm{Ci}\). What is its activity in becquerels?

Cobalt-60 has a half-life of \(5.26\) years. If \(1.00 \mathrm{~g}\) of \({ }^{60} \mathrm{Co}\) were allowed to decay, how many grams would be left after (a) one half-life? (b) two half-lives? (c) four half-lives? (d) ten half-lives?

The half-life of I-123 is 13 hours. If \(10 \mathrm{mg}\) of I-123 are administered to a patient, how much I-123 remains after 3 days and 6 hours?

When the nuclide \({ }_{98}^{249} \mathrm{Cf}\) was bombarded with \({ }_{7}^{15} \mathrm{~N}\), four neutrons and a new transuranium element were formed. Write the nuclear equation for this transmutation.

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