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Chapter 45: Problem 47

Complete the following equations. (See Appendix \(\mathrm{H}\) for a table of the elements.) (a) \({ }_{11}^{23} \mathrm{Na}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }^{26} \mathrm{Mg}+\) ? (d) \({ }_{5}^{10} \mathrm{~B}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{6}^{13} \mathrm{~N}+?\) (b) \({ }_{29}^{64} \mathrm{Cu} \rightarrow{ }_{+1}^{0} e+?\) (e) \({ }_{48}^{105} \mathrm{Cd}+{ }_{-1}^{0} e=\rightarrow ?\) (c) \({ }^{106} \mathrm{Ag} \rightarrow{ }^{106} \mathrm{Cd}+\) ? (f) \({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{90}^{234} \mathrm{Th}+\) ?

Short Answer

Expert verified
(a) \( \,_{1}^{1} \mathrm{H} \), (d) \( \,_{1}^{1} \mathrm{H} \), (b) \( \,_{28}^{64} \mathrm{Ni} \), (e) \( \,_{47}^{105} \mathrm{Ag} \), (c) \( \,_{-1}^{0} e \,\), (f) \( \,_{2}^{4} \mathrm{He} \).

Step by step solution

01

Reaction (a) Analysis

First, balance the mass numbers and atomic numbers for the equation \( \,_{11}^{23} \mathrm{Na}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }^{26} \mathrm{Mg}+\,? \,\). Here, the sum of mass numbers on the left side is \(23 + 4 = 27\). On the right, \(26 + x = 27\), which gives \(x = 1\) for the missing particle. The sum of atomic numbers is \(11 + 2 = 13\) on the left, and \(12 + y\) on the right, giving \(y = 1\). Thus, \( \,_{1}^{1} \mathrm{H}\,\) is the missing particle.
02

Reaction (b) Analysis

For \( \,_{29}^{64} \mathrm{Cu} \rightarrow{ }_{+1}^{0} e+?\,\), this is a beta decay, where a neutron turns into a proton. Therefore, the mass number remains the same, \(64\). The atomic number becomes \(29 - 1 = 28\), so the resulting element is \( \,_{28}^{64} \mathrm{Ni}\,\).
03

Reaction (c) Analysis

Given \( \,^{106} \mathrm{Ag} \rightarrow{ }^{106} \mathrm{Cd}+? \,\), a beta decay is likely happening, where a neutron converts to a proton and an electron (beta particle) is emitted. The mass number stays \(106\), and the atomic transition suggests an electron emission, which confirms \( \,_{-1}^{0} e \,\) is the emitted particle.
04

Reaction (d) Analysis

For \( \,_{5}^{10} \mathrm{~B}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{6}^{13} \mathrm{~N}+?\,\), the mass numbers sum \(10 + 4 = 14\) on the left. On the right \(13 + x = 14\), leading to \(x = 1\). The sum of atomic numbers \(5 + 2 = 7\), on the right \(6 + y\), yielding \(y = 1\). The missing particle is \( \,_{1}^{1} \mathrm{H} \,\).
05

Reaction (e) Analysis

With \( \,_{48}^{105} \mathrm{Cd}+{ }_{-1}^{0} e= \rightarrow ?\,\), a beta capture occurs. The atomic number reduces by one as an electron combines, leading to \(Z = 47\), while the mass number remains \(105\). Thus, the resulting element is \( \,_{47}^{105} \mathrm{Ag}\,\).
06

Reaction (f) Analysis

In the equation \( \,_{92}^{238} \mathrm{U}\rightarrow{ }_{90}^{234} \mathrm{Th}+?\,\), this is an alpha decay, emitting an \( \,_{2}^{4} \mathrm{He}\). The sum of mass numbers on both sides \(238\) is verified \(234 + 4 = 238\), and similarly for atomic numbers \(92\), validated \(90 + 2 = 92\). Hence, the emitted particle is \( \,_{2}^{4} \mathrm{He}\).

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Key Concepts

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

Alpha Decay
Alpha decay is a type of nuclear reaction where an unstable nucleus emits an alpha particle. An alpha particle is composed of 2 protons and 2 neutrons, making it identical to a helium-4 nucleus. This emission reduces the original element's atomic number by 2 and the mass number by 4. For example, in the reaction \( \ _{92}^{238} \mathrm{U} \rightarrow \ _{90}^{234} \mathrm{Th}+\ _{2}^{4} \mathrm{He} \), uranium-238 decays into thorium-234 by emitting an alpha particle.
  • The atomic numbers are balanced: \(92 = 90 + 2\).
  • The mass numbers are also balanced: \(238 = 234 + 4\).
Alpha particles have a relatively high energy and they can be stopped by a sheet of paper or even human skin, offering a degree of protection from this type of radiation.
Beta Decay
Beta decay is another form of nuclear decay where a neutron in an atom's nucleus is transformed into a proton and a beta particle, which is an electron or positron. A famous example of beta decay is \( \ _{29}^{64} \mathrm{Cu} \rightarrow \ _{28}^{64} \mathrm{Ni} + \ _{+1}^{0} e\).
  • Matter conversion: Neutron converts into proton.
  • Radiation emission: Electron (beta particle) is emitted.
  • Balance of atomic numbers: Original becomes 29, new is 28 (the added beta particle balances this).
  • Constant mass number: 64 stays the same.
Beta decay plays a crucial role in the transformation of elements and helps to create new elements with different properties. It emits radiation capable of penetrating human tissue, though typically can be stopped by a sheet of metal or thick plastic.
Balancing Nuclear Equations
Balancing nuclear equations involves ensuring that the sum of atomic numbers and mass numbers are equal on both sides of the equation. This ensures that matter is conserved during the reaction. Let's look at the reaction \( \ _{5}^{10} \mathrm{~B} + \ _{2}^{4} \mathrm{He} \rightarrow \ _{6}^{13} \mathrm{~N} + \ _{1}^{1} \mathrm{H} \).
  • Sum of mass numbers on the left: \(10 + 4 = 14\).
  • Sum of mass numbers on the right: \(13 + 1 = 14\).
  • Sum of atomic numbers on the left: \(5 + 2 = 7\).
  • Sum of atomic numbers on the right: \(6 + 1 = 7\).
This type of exercise is crucial for understanding how particles are conserved in reactions, ensuring equations correctly represent physical processes.
Element Symbols
Element symbols are shorthand notations used to quickly identify chemical elements in nuclear equations. They typically consist of one or two letters, often derived from the Latin or English names of the elements. The symbol is accompanied by two numbers:
  • The atomic number (bottom left) indicates the number of protons. It defines the element.
  • The mass number (upper left) is the sum of protons and neutrons.For example, in \( \ _{29}^{64} \mathrm{Cu}\):
  • "Cu" is the symbol for copper.
  • 29 is the atomic number, showing it has 29 protons.
  • 64 is the mass number, meaning it has 64 protons and neutrons in total.
Understanding element symbols is vital for interpreting and writing nuclear equations effectively. This knowledge helps predict the properties and reactivity of different elements in nuclear reactions.

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Most popular questions from this chapter

How many protons, neutrons, and electrons are there in \((a){ }^{3} \mathrm{He}\), (b) \(^{12} \mathrm{C}\), and \((c)^{206} \mathrm{~Pb}\) ? (a) The atomic number of He is 2; therefore, the nucleus must contain 2 protons. Since the mass number of this isotope is 3 , the sum of the protons and neutrons in the nucleus must equal 3; therefore, there is 1 neutron. The number of electrons in the atom is the same as the atomic number, 2 . (b) The atomic number of carbon is 6 ; hence, the nucleus must contain 6 protons. The number of neutrons in the nucleus is equal to \(12-6=6\). The number of electrons is the same as the atomic number, 6 . (c) The atomic number of lead is 82 ; hence, there are 82 protons in the nucleus and 82 electrons in the atom. The number of neutrons is \(206-82=124\).

The half-life of uranium-238 is about \(4.5 \times 10^{9}\) years, and its end product is lead-206. We notice that the oldest uraniumbearing rocks on Earth contain about a \(50: 50\) mixture of \({ }^{238} \mathrm{U}\) and \({ }^{206} \mathrm{~Pb}\). Roughly, what is the age of these rocks? Apparently about half the \({ }^{238} \mathrm{U}\) has decayed to \({ }^{206} \mathrm{~Pb}\) during the existence of the rock. Hence, the rock must have been formed about \(4.5\) billion years ago.

By natural radioactivity \({ }^{238} \mathrm{U}\) emits an \(\alpha\) -particle. The heavy residual nucleus is called \(\mathrm{UX}_{1} . \mathrm{UX}_{1}\) in turn emits a beta particle. The resultant nucleus is called \(\mathrm{UX}_{2} .\) Determine the atomic number and mass number for \((a) \mathrm{UX}_{1}\) and \((b) \mathrm{UX}_{2}\).

Uranium-238 \(\left({ }_{92}^{238} \mathrm{U}\right)\) ) is radioactive and decays into a succession of different elements. The following particles are emitted before the nucleus reaches a stable form: \(\alpha, \beta, \beta, \alpha, \alpha, \alpha, \alpha, \alpha, \beta, \beta, \alpha, \beta\), \(\beta\), and \(\alpha\left(\beta\right.\) stands for "beta particle, \(" e^{-}\) ). What is the final stable nucleus? The original nucleus emitted 8 alpha particles and 6 beta particles. When an alpha particle is emitted, \(Z\) decreases by 2 , since the alpha particle carries away a charge of \(+2 e\). A beta particle carries away a charge of \(-1 e\), and so as a result the charge on the nucleus must increase to \((Z+1) e\). We then have, for the final nucleus, Final \(Z=92+6-(2)(8)=82\) Final \(A=238-(6)(0)-(8)(4)=206\) The final stable nucleus is \({ }_{82}^{206} \mathrm{~Pb}\) -

The half-life of carbon-14 is \(5.7 \times 10^{3}\) years. What fraction of a sample of \({ }^{14} \mathrm{C}\) will remain unchanged after a period of five halflives?
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