Problem 43
Question
The number of \(\alpha\) and \(\beta\) particle emitted in the nuclear reaction \({ }^{228} \mathrm{Th}_{90} \longrightarrow{ }^{212} \mathrm{Bi}_{83}\) are (a) \(4 \alpha\) and \(1 \beta\) (b) \(3 \alpha\) and \(7 \beta\) (c) \(8 \alpha\) and \(1 \beta\) (d) \(4 \alpha\) and \(7 \beta\)
Step-by-Step Solution
Verified Answer
The answer is (a) 4 alpha and 1 beta.
1Step 1: Understand the Process
In nuclear reactions, an alpha (\(\alpha\)) particle consists of 2 protons and 2 neutrons, which decreases the atomic number by 2 and the mass number by 4. A beta (\(\beta\)) particle changes a neutron into a proton, increasing the atomic number by 1 while keeping the mass number constant.
2Step 2: Analyze the Original and Final Nuclei
The original nucleus is thorium-228 (\(^{228} \mathrm{Th}_{90}\)), and it decays into bismuth-212 (\(^{212} \mathrm{Bi}_{83}\)). This means the mass number decreases from 228 to 212 and the atomic number decreases from 90 to 83.
3Step 3: Calculate Number of Alpha Particles
Each alpha particle reduces the mass number by 4. The change in mass number is 228 - 212 = 16. Therefore, the number of alpha particles emitted is \(\frac{16}{4} = 4\).
4Step 4: Calculate Number of Beta Particles
Alpha decay has already accounted for a decrease of 8 in the atomic number (90 to 82). The final atomic number is 83, which means one beta particle must be emitted to increase the atomic number from 82 (after alpha decay) to 83 (final). Thus, one beta particle is emitted.
Key Concepts
Alpha DecayBeta DecayMass Number Change
Alpha Decay
Alpha decay is a type of radioactive decay where a nucleus emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons. This particle is essentially a helium nucleus, represented as \(^{4}He_{2}\).When a nucleus undergoes alpha decay, it loses these 4 nucleons, which affects both the atomic number and the mass number:
- The atomic number decreases by 2, as 2 protons are lost.
- The mass number decreases by 4, because of the total loss of 4 nucleons.
Beta Decay
Beta decay involves the transformation of a neutron into a proton within a nucleus, accompanied by the emission of a beta particle (an electron) and an antineutrino. In this process:
- The atomic number increases by 1, as a new proton is formed.
- The mass number remains unchanged because neither protons nor neutrons are lost or gained overall.
Mass Number Change
The mass number of an atom, represented by \(A\), is the sum of the number of protons and neutrons in its nucleus. It is an important quantity as it reflects the weight of the nucleus, and when nuclear reactions occur, changes in the mass number can indicate the type and amount of particles emitted.During alpha decay, the mass number decreases because 4 nucleons (2 protons and 2 neutrons) are expelled from the nucleus. In the example given, thorium's mass number decreases from 228 to 212. The change in mass number, computed as \(228 - 212 = 16\), shows us that 4 alpha particles have been emitted since each alpha emission reduces the mass number by 4.Notably, beta decay does not affect the mass number. This is because while a neutron is transformed into a proton, the total number of nucleons remains unchanged. Therefore, mass number changes most significantly through processes involving the expulsion or capture of nucleons, such as alpha decay. This understanding allows students to track and analyze nuclear reactions by observing how the mass number varies.
Other exercises in this chapter
Problem 41
When a radioactive element emits successively one \(\alpha\) particle and two \(\beta\) particles, the mass number of the daughter element (a) is reduced by 4 u
View solution Problem 42
A nuclide of an alkaline earth metal undergoes radioactive decay by emission of the \(\alpha\) particle in succession. The group of the periodic table to which
View solution Problem 44
The compound used in enrichment of the uranium in nuclear power plant is (a) \(\mathrm{UF}_{6}\) (b) \(\mathrm{U}_{3} \mathrm{O}_{8}\) (c) \(\mathrm{UCl}_{4}\)
View solution Problem 45
\({ }_{2} \mathrm{U}^{238}\) is radioactive and it emits \(\alpha\) and \(\beta\) particles to form \({ }_{82} \mathrm{~Pb}^{206} .\) The number of \(\alpha\) a
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