Problem 26
Question
In successive emission of \(\beta\) and \(\alpha\) particles, how many \(\alpha\) and \(\beta\) particles should be emitted for the natural \(\left(4 n+1\right.\) series) conversion of \({ }_{94} \mathrm{Pu}^{241}\) to \({ }_{92} \mathrm{U}^{233}\) ? (a) \(2 \alpha, \beta\) (b) \(2 \alpha, 2 \beta\) (c) \(2 \alpha, 3 \beta\) (d) \(\alpha, 2 \beta\)
Step-by-Step Solution
Verified Answer
(b) 2 \(\alpha\), 2 \(\beta\)
1Step 1: Identifying Initial and Final Nuclei
We start with \({}_{94}\mathrm{Pu}^{241}\) and want to convert it to \({}_{92}\mathrm{U}^{233}\). indicates the plutonium to uranium conversion through nuclear decay processes that involve the emission of alpha (\(\alpha\)) and beta (\(\beta\)) particles.
2Step 2: Understanding Particle Emission Results
An alpha particle emission reduces the atomic number by 2 and the mass number by 4. A beta particle emission increases the atomic number by 1 without changing the mass number.
3Step 3: Calculate Alpha Emissions Needed
To go from a mass number of 241 to 233, we calculate the difference: \( 241 - 233 = 8 \). Each alpha particle reduces the mass number by 4, so: \(\frac{8}{4} = 2\alpha\).
4Step 4: Determine Change in Atomic Number
Going from an atomic number of 94 to 92 means reducing the atomic number by 2. An alpha emission reduces the atomic number by 2. Two alpha emissions therefore lower the atomic number by 4, so intermediate state would be \({}_{90}X^{233}\), two atomic numbers short of uranium.
5Step 5: Calculate Beta Emissions Needed
Since two alpha emissions reduce the atomic number from 94 to 90, \(2\beta\) emissions are needed to raise the intermediate atomic number from 90 to 92, completing the conversion to uranium.
6Step 6: Confirm Correct Option
According to the calculated emissions, \(2\alpha\) and \(2\beta\) particles are needed for the natural conversion of \({}_{94}\mathrm{Pu}^{241}\) to \({}_{92}\mathrm{U}^{233}\). The answer is option (b).
Key Concepts
Alpha EmissionBeta EmissionAtomic NumberMass Number
Alpha Emission
Alpha emission is one of the processes of nuclear decay where an unstable nucleus loses energy by emitting an alpha particle. An alpha particle is made up of two protons and two neutrons, identical to a helium nucleus.
When an atom undergoes alpha emission, its atomic number decreases by 2 and its mass number decreases by 4.
This means the atom loses some of its protons and neutrons, becoming a different element.
For example, when a _{94}^{241} ext{Pu} atom emits an alpha particle, it transforms into a new element with an atomic number of _{92} and a mass number of ^{237} .
When an atom undergoes alpha emission, its atomic number decreases by 2 and its mass number decreases by 4.
This means the atom loses some of its protons and neutrons, becoming a different element.
For example, when a _{94}^{241} ext{Pu} atom emits an alpha particle, it transforms into a new element with an atomic number of _{92} and a mass number of ^{237} .
Beta Emission
Beta emission is another form of nuclear decay where a beta particle is emitted from the nucleus of an atom. Beta particles can be either electrons or positrons, although in beta decay related to this exercise, electrons are typically involved.
During beta emission, a neutron in the nucleus is converted into a proton, releasing an electron in the process.
During beta emission, a neutron in the nucleus is converted into a proton, releasing an electron in the process.
- This conversion increases the atomic number of the atom by 1, as the number of protons in the nucleus increases.
- However, the mass number remains unchanged because a neutron is replaced by a proton.
Atomic Number
The atomic number is fundamental in identifying a chemical element. It represents the number of protons in the nucleus of an atom. The atomic number is what makes each element unique because no two elements have the same atomic number.
Alpha and beta emissions can alter the atomic number of an atom:
Alpha and beta emissions can alter the atomic number of an atom:
- An alpha emission decreases the atomic number by 2.
- A beta emission increases the atomic number by 1.
Mass Number
The mass number of an atom is the total number of protons and neutrons in its nucleus. This is different from the atomic number, which only counts protons.
While the mass number doesn't identify the element (that's the job of the atomic number), it tells you the isotope of an element.
Alpha emissions impact the mass number directly by reducing it due to the loss of two protons and two neutrons, totaling four.
While the mass number doesn't identify the element (that's the job of the atomic number), it tells you the isotope of an element.
Alpha emissions impact the mass number directly by reducing it due to the loss of two protons and two neutrons, totaling four.
- In our conversion example, alpha emissions decrease the mass number from ^{241} to ^{233} .
- Beta emissions do not alter the mass number since they involve a neutron converting to a proton, which neither adds nor removes mass in the form of particles.
Other exercises in this chapter
Problem 24
The half lives of two radioactive nuclides \(\mathrm{A}\) and \(\mathrm{B}\) are 1 and 2 min respectively. Equal weights of \(A\) and \(B\) are taken separately
View solution Problem 25
In the nuclear fusion \({ }_{1} \mathrm{H}^{2}+{ }_{1} \mathrm{H}^{2} \longrightarrow{ }_{2} \mathrm{He}^{4}\), the masses of \({ }_{1} \mathrm{H}^{2}\) and \({
View solution Problem 27
What will be the binding energy of \(\mathrm{O}^{16}\), if the mass defect is \(0.210\) amu? (a) \(1.89 \times 10^{10} \mathrm{~J} \mathrm{~mol}^{-1}\) (b) \(1.
View solution Problem 28
The compound used for the preparation of \(U F_{6}\) in the enrichment of \({ }_{92} \mathrm{U}^{235}\) is (a) HF (b) \(\mathrm{CaF}_{2}\) (c) \(\mathrm{ClF}_{3
View solution