Problem 64
Question
Write the balanced nuclear equation for each of the following nuclear reactions: (a) electron capture by iron-55, (b) beta emission by potassium-42, (c) positron emission by ruthenium-93, (d) alpha emission by californium- 251 .
Step-by-Step Solution
Verified Answer
\(^{55}_{26}Fe + e^- \rightarrow ^{55}_{25}Mn + u_e\), \(^{42}_{19}K \rightarrow ^{42}_{20}Ca + e^- + \bar{u}_e\), \(^{93}_{44}Ru \rightarrow ^{93}_{43}Tc + e^+ + u_e\), \(^{251}_{98}Cf \rightarrow ^{247}_{96}Cm + ^{4}_{2}He\).
1Step 1: Understanding Electron Capture
Electron capture is a process in which an inner shell electron combines with a proton in the nucleus to form a neutron and emit an electron neutrino. The atomic number decreases by one, but the mass number remains the same. In Iron-55, this can be represented as \(e^- + ^{55}_{26}Fe \rightarrow ^{55}_{25}Mn + u_e\).
2Step 2: Writing the Beta Emission Equation
Beta emission means a neutron in the nucleus is converted to a proton and emits a beta particle (electron) and an antineutrino. The atomic number increases by one while the mass number remains constant. For Potassium-42, the equation is \(^{42}_{19}K \rightarrow ^{42}_{20}Ca + e^- + \bar{u}_e\).
3Step 3: Describing Positron Emission
Positron emission occurs when a proton is transformed into a neutron with the emission of a positron and a neutrino. The atomic number decreases by one, but the mass number remains unchanged. The equation for Ruthenium-93 is \(^{93}_{44}Ru \rightarrow ^{93}_{43}Tc + e^+ + u_e\).
4Step 4: Alpha Emission Representation
Alpha emission involves the release of an alpha particle from the nucleus, which contains two protons and two neutrons, reducing the atomic number by two and the mass number by four. Californium-251 undergoes alpha emission as follows: \(^{251}_{98}Cf \rightarrow ^{247}_{96}Cm + ^{4}_{2}He\), where \(^{4}_{2}He\) represents the alpha particle.
Key Concepts
Understanding Electron CaptureBeta Emission in NucleiPositron Emission and IsotopesAlpha Emission and Atomic Changes
Understanding Electron Capture
Electron capture is a fascinating process that occurs inside an atom's nucleus. Atomic nuclei are made of protons and neutrons, with protons being positively charged and neutrons electrically neutral. Sometimes, an atom's nucleus captures one of its own electrons, typically from the innermost shell.
This captured electron combines with a proton to produce a neutron and a neutrino, a nearly massless particle. What's spectacular about electron capture is that it causes the atom to transform into a different element because the number of protons, which determines the element's identity, decreases by one. However, the mass number, which is the total number of protons and neutrons, doesn't change.
For example, in the electron capture by iron-55, one electron (\(e^-\)) is captured by the nucleus (\(^{55}_{26}Fe\)), resulting in a manganese-55 atom (\(^{55}_{25}Mn\)) and the emission of an electron neutrino (\(u_e\)).
This captured electron combines with a proton to produce a neutron and a neutrino, a nearly massless particle. What's spectacular about electron capture is that it causes the atom to transform into a different element because the number of protons, which determines the element's identity, decreases by one. However, the mass number, which is the total number of protons and neutrons, doesn't change.
For example, in the electron capture by iron-55, one electron (\(e^-\)) is captured by the nucleus (\(^{55}_{26}Fe\)), resulting in a manganese-55 atom (\(^{55}_{25}Mn\)) and the emission of an electron neutrino (\(u_e\)).
Beta Emission in Nuclei
Beta emission is another type of nuclear decay which plays a key role in the natural transmutation of elements. Unlike electron capture, beta emission increases the atomic number of an element by one, while keeping the mass number unchanged. This increase is due to the conversion of a neutron in the nucleus into a proton, which is accompanied by the release of a beta particle (which is an electron) and an antineutrino.
Consider potassium-42 (\(^{42}_{19}K\)). It emits a beta particle (\(e^-\)), along with an antineutrino (\(\bar{u}_e\)), resulting in calcium-42 (\(^{42}_{20}Ca\)). This process is a crucial part of many natural radioactive series and has applications in medical imaging and treatment, as well as carbon dating in archaeology.
Consider potassium-42 (\(^{42}_{19}K\)). It emits a beta particle (\(e^-\)), along with an antineutrino (\(\bar{u}_e\)), resulting in calcium-42 (\(^{42}_{20}Ca\)). This process is a crucial part of many natural radioactive series and has applications in medical imaging and treatment, as well as carbon dating in archaeology.
Positron Emission and Isotopes
Positron emission is quite the opposite of beta emission. During positron emission, a proton inside the nucleus is transformed into a neutron and releases a positron (\(e^+\)), which is the antimatter counterpart of the electron, along with a neutrino. This nuclear reaction, again, leaves the mass number constant but lowers the atomic number by one, leading to the transformation into a new element.
For instance, ruthenium-93 (\(^{93}_{44}Ru\)) undergoes positron emission to become technetium-93 (\(^{93}_{43}Tc\)). Positron emission is not just a theoretical concept but is employed in positron emission tomography (PET), a cutting-edge medical imaging technique that helps detect diseases in the body.
For instance, ruthenium-93 (\(^{93}_{44}Ru\)) undergoes positron emission to become technetium-93 (\(^{93}_{43}Tc\)). Positron emission is not just a theoretical concept but is employed in positron emission tomography (PET), a cutting-edge medical imaging technique that helps detect diseases in the body.
Alpha Emission and Atomic Changes
Alpha emission is another way that unstable nuclei can reach a more stable state. An alpha particle, which consists of two protons and two neutrons bound together (\(^{4}_{2}He\)), is ejected from the nucleus. This emission results in the reduction of both atomic and mass numbers—by two and four, respectively. Alpha particles are relatively heavy and carry a double positive charge.
Californium-251 (\(^{251}_{98}Cf\) demonstrates alpha emission by releasing an alpha particle to form curium-247 (\(^{247}_{96}Cm\)). This process is crucial for understanding the synthesis and decay of heavy elements and is a common type of radiation in radioactive decay series seen in elements such as uranium and radium.
Californium-251 (\(^{251}_{98}Cf\) demonstrates alpha emission by releasing an alpha particle to form curium-247 (\(^{247}_{96}Cm\)). This process is crucial for understanding the synthesis and decay of heavy elements and is a common type of radiation in radioactive decay series seen in elements such as uranium and radium.
Other exercises in this chapter
Problem 54
Calculate the energy equivalent in \(\mathrm{kJ}\) of \(1.00 \mathrm{~g}\) of mass.
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View solution Problem 65
Write the symbols, including the atomic and mass numbers, for the radionuclides that would give each of the following products: (a) fermium- 257 by alpha emissi
View solution Problem 66
Each of the following nuclides forms by the decay mode described. Write the symbols of the parents, giving both atomic and mass numbers: (a) rubidium- 80 formed
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