Chapter 11: Atomic and nuclear structure (C2307838)

 #toc { border: 1px solid #bba; background-color: #f7f8ff; padding: 1em; font-size: 90%; text-align: center; } #toc-header { display: inline; padding: 0; font-size: 100%; font-weight: bold; } #toc ul { list-style-type: none; margin-left: 0; padding-left: 0; text-align: left; } .toc2 { margin-left: 1em; } .toc3 { margin-left: 2em; } .toc4 { margin-left: 3em; }Table of Contents Last modified: 2449d agoWord count: 1,584 wordsLegend: Key principles // Storyline

1 Atomic spectra

The atomic spectrum includes either:

• Emission line, which are discrete/quantized [atomic] energy levels which cause a photon to emit. There are thus a few colored lines, overlaid on a black background
• Absorption line, which are wavelengths where photons are absorbed, are missing. This thus causes a colored background, overlaid by discrete lines in black

 Formative learning activity Maps to RK11.A What is the atomic spectra?

2 Atomic nucleus

The atomic nucleus is the dense region in the center of an atom, consisting of protons and neutrons. Protons are positively charged, and neutrons are neutrally charged. Protons and neutrons are essentially equal in mass.

Radioactive decay is the process by which an unstable atom spontaneously disintegrates, and emits with it ionizing particles. Half-life is the time taken for a quantity to fall to half its value. Half-life is probabilistic, in that it cannot determine whether a single atom will decay or not, but if analyzed globally, will show that half the substance has decayed (to something else) within that half-life. The half-life is given by $t_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$, where $ln(2)\approx 0.693$, and $\lambda$ is the decay constant of a substance.

“For example, if the half-life of a substance if 5 years, if we begin with 200g, how much substance will there be in 20 years?” Jamie asked.

“After 5 years there will be 100g, after 10 years there will be 50 grams, after 15 years there will be 25 grams, and after 20 years there will be 12.5 grams,” Mandy answered.

“Another example,” Jamie continued, “if the initial quantity is 100g, the current quantity is 25g, and the time taken is 10 years, what is the half-life?”

“Going from 100 to 50 to 25, there have been 2 half-lives,” Mandy thought, “This means each half life has taken 5 years .”

The different types of radioactive decay include:

• Alpha decay, which is decay that occurs with the emission of an alpha particle. Alpha particle is a $He$ nucleus, which is 2 protons and 2 neutrons, namely, $\alpha=^{4}_{2}He$. Because an alpha particle is emitted, the resulting decay can be discovered by the difference in mass and atomic numbers. For example, in the alpha decay of Uranium-238, the formula $\ce{^{238}_{92}U -> ^{a}_{b}X + ^{4}_{2}He}$ can be drawn. Given that $238=\alpha +4$ and $92=b+2$ for the mass and atomic numbers respectively, therefore, reshuffling, $\alpha=234$ and $b=90$. The element with the atomic number 90 is Thorium Th. Therefore, $\ce{^{238}_{92}U -> ^{234}_{90}Th + ^{4}_{2}He}$.
• Beta decay, which is decay that occurs with the emission of a beta particle. It occurs either because a nucleus contains too many neutrons or protons, thereby converting one to another or vice versa. It includes:
• Beta-minus decay, which is where excess neutrons are converted into protons. The relevant beta particle in beta-minus decay is an electron ($^0_{-1}e^-$). An anti-neutrino ($^0_0\overline{v}$) is also emitted. When drawing up an equation, it is important to ensure conservation of charge, such that if an electron is emitted, it must be balanced by the emission of a proton too. The net equation is $^1_{0}n -> ^1_1 p^+ + ^0_{-1}e^- + ^0_0\overline{v}$. For example, in the beta-minus decay of Thorium-234, the formula $\ce{^{234}_{90}Th -> ^a_b X + ^0_{-1}e^- + ^0_0\overline{v}}$ can be drawn. Thus, $234=a+0+0$ and $90=b-1+0$, meaning $a=234$ and $b=91$. The element with the atomic number 91 is Protactinium Pa. Thus, the formula becomes $\ce{^{234}_{90}Th -> ^{234}_{91}Pa + ^0_{-1}e^- + ^0_0\overline{v}}$
• Beta-plus decay, which is where excess protons are converted into neutrons. The relevant beta particle in beta-plus decay is a positron. A positron ($^0_{+1}e^+$) is the positive counterpart of an electron (remember protons are heavier than an electron, though they have exact and opposite charge, and so therefore aren’t the exact opposite of an electron). A neutrino ($^0_{0} v$) is also emitted. The net equation is $\ce{^1_{1}p^+ -> ^1_{0}n + ^0_{+1}e^+ + ^0_{0}v}$. For example, in the beta-plus decay of Sodium-22, the formula $\ce{^22_{11}Na -> ^a_{b}X + ^0_{+1}e^+ + ^0_{0}v}$ can be drawn. Thus, $22=a+0+0$ and $11=b+1+0$, meaning $a=22$ and $b=10$. The element with the atomic number 10 is Neon Ne. Thus, the formula becomes $\ce{^{22}_{11}Na -> ^22_{10}Ne + ^0_{+1}e^+ + ^0_{0} v}$
• Electron capture, which is where an electron is absorbed by a proton to form a neutron. The net equation is $\ce{^1_{1}p^+ + ^0_{-1}e^- -> ^1_{0}n + ^0_{0}v}$, which is similar to beta-plus decay. For example, in the electron capture of Mercury-201, the formula $\ce{^{201}_{80}Hg + ^0_{-1}e^- -> ^a_{b}X + ^1_{0}n + ^0_{0}v}$ can be drawn. Thus, $201+0=a+1+0$ and $80-1=b+0+0$, meaning $a=200$ and $b=79$. The element with the atomic number 79 is Gold Au. Thus, the formula becomes $\ce{^{201}_{80}Hg + ^0_{-1}e^- -> ^{201}_{79}Au + ^0_{0}v}$
• Gamma decay, which is where a nucleus may be excited, as a result of emitting an $\alpha$ or $\beta$ particle. To stabilize, it can emit the energy in the form of a gamma ray. Gamma ray is pure energy, and thus has neither mass nor charge. For example, electron-positron annihilation is when an electron ($^0_{-1}e^-$) and positron ($^0_{+1}e^+$) collide, expressed in the formula $\ce{^0_{-1}e^- + ^0_{+1}e^+ -> ^0_{0}\gamma + ^0_{0}\gamma}$

Mass-energy equivalence is $E=m.c^2$, where $m$ is the change in mass, and $c$ is the speed of light $3\times 10^8 m/s$, and shows energy content of a mass. It can determine how much energy is created when masses disappear, for example, in the case of electron-positron annihilation. Also, it can determine the nuclear binding energy, which is the energy required to split a nucleus into its components. In a nucleus, it is notable that the mass of the nucleus, is less than the mass of the constituent components. This is due to mass defect/deficit, which is that binding energy is required to keep the components together, thereby requiring some mass to be converted into this energy. This mass defect can be used to find the energy equivalence, which is the nuclear binding energy.

Nuclear fusion is the combining of two light nuclei to form a more stable heavier nucleus. Nuclear fission is the splitting of a heavier nucleus to form more stable lighter nuclei. Nuclei heavier than iron-56tend to undergo fission, and nuclei lighter than iron-56 tend to undergo fusion. Both fusion and fission release energy. Noting that creating bonds releases energy (see ), and breaking bonds requires energy, although it is evident why fusion releases energy, it becomes unclear why fission releases energy. Although fission initially requires energy, the energy released in subsequence is much larger, meaning fission net releases energy.

 Formative learning activity Maps to RK11.B What does the atomic nucleus consist of?

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