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Chapter 33: Problem 41

The number of \(\beta\) -particles, if a radioactive element \(90^{238}\) decays into \({ }_{83} Y^{222}\) is : (a) 4 (b) 6 (c) 2 (d) 1

Short Answer

Expert verified
The number of \( \beta \)-particles emitted is 1.

Step by step solution

01

Analyze the given decay process

In the given problem, the radioactive element starts as \( {}^{238}_{90}X \) and decays to \( {}^{222}_{83}Y \). Here, \( X \) is the parent isotope, and \( Y \) is the daughter isotope.
02

Define the decay equations

During radioactive decay, the nucleus can undergo alpha or beta decay: \( \alpha \)-particles have a mass number of 4 and an atomic number of 2, while \( \beta \)-particles do not change the mass number but change the atomic number by \( +1 \).
03

Calculate mass number changes

Start with the mass number: \( 238 \to 222 \). The loss in mass number is \( 238 - 222 = 16 \). Each \( \alpha \)-particle decreases the mass number by 4, therefore number of \( \alpha \)-particles emitted: \( \frac{16}{4} = 4 \).
04

Calculate atomic number changes

The initial atomic number is 90 and the final is 83. The change in atomic number is \( 90 - 83 = 7 \). Given that each \( \alpha \)-particle decreases the atomic number by 2, hence loss due to 4 \( \alpha \)-particles: \( 4 \times 2 = 8 \).
05

Determine the number of beta particles

After accounting for reduction due to \( \alpha \)-particles, the change in atomic number needing adjustment by \( \beta \)-particles is \( 7 - 8 = -1 \). As each \( \beta \)-particle increases the atomic number by 1, 1 \( \beta \)-particle is emitted to adjust for this.

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Key Concepts

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

Alpha Decay
In alpha decay, a radioactive nucleus emits an alpha particle, which is identical to a helium nucleus. It consists of two protons and two neutrons. This accounts for a reduction in mass number by 4 and a decrease in the atomic number by 2.
When a nucleus like \( {}^{238}_{90}X \) undergoes alpha decay, the transformation leads to a new element with reduced atomic number and mass number.
  • Example: The emission of an alpha particle from \( {}^{238}_{90}X \) reduces its mass number to 234 and atomic number to 88.
  • As a result: \({}^{234}_{88}Y\) is formed.
Alpha decay is common among heavy elements like uranium and radium.
It is a type of decay that especially stabilizes large nuclei that may have too many protons clustered together.
Beta Decay
Beta decay is an interesting process where a neutron in the nucleus turns into a proton, or vice versa, emitting a beta particle in the process. This beta particle is simply a high-energy, high-speed electron or positron.
In the exercise, beta decay increases the atomic number by 1 while the mass number remains unchanged.
  • Important Note: In beta-minus decay, a neutron turns into a proton, thereby increasing the atomic number.
  • For beta-plus decay, a proton turns into a neutron, resulting in a decrease of the atomic number.
In the solving process, our radioactive element needed a single beta emission to boost the atomic number by one, correcting the change after the emission of multiple alpha particles.
The beta decay process ensured that the final atomic number was as specified in the exercise, \( {}_{83}Y^{222} \).
Mass and Atomic Number Calculations
Understanding the calculations for mass and atomic numbers is crucial in nuclear decay problems. Each decay mode affects these numbers differently.In the provided exercise:
  • The initial mass number was 238, while the final number is 222. The difference, 16, results from the emission of four alpha particles \( (4 \times 4 = 16) \).
  • The atomic number changed from 90 to 83. The emission of four alpha particles leads to a reduction by 8 \( (4 \times 2 = 8) \), resulting in a need to adjust back by +1 via one beta particle emission \((+1)\).
To analyze such decays, recognize how alpha decay subtracts from both mass and atomic numbers, whereas beta decay adjusts mainly the atomic number only. By understanding these nuclear arithmetic rules, you can determine quantities like the number of alpha or beta particles released.
Practicing such calculations builds a strong conceptual framework for tackling more complex nuclear decay problems.

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Most popular questions from this chapter

The rest mass energy of electron is : (a) \(0.8 \mathrm{MeV}\) (b) \(1.66 \mathrm{amu}\) (c) \(0.5119 \mathrm{MeV}\) (d) none of these

What is the binding energy per nucleon of \({ }_{6} \mathrm{C}^{12}\) nucleus? Given: mass of \(C^{12}\left(m_{c}\right)=12.000 \mathrm{u}\) mass of proton \(\left(m_{p}\right)=1.0078 \mathrm{u}\) mass of neutron \(\left(m_{n}\right)=1.0087 \mathrm{u}\) and \(1 \mathrm{amu}=931.4 \mathrm{MeV}\) (a) \(5.26 \mathrm{MeV}\) (b) \(10.11 \mathrm{MeV}\) (c) \(15.65 \mathrm{MeV}\) (d) \(7.68 \mathrm{MeV}\)

What should be the activity of a radioactive sample of mass \(m\) and decay constant \(\lambda\), after time \(t ?\) Take molecular weight of the sample be \(M\) and the Avagadro number be \(N_{A}:\) (a) \(A:=\left(\frac{M i N_{A} \lambda}{m}\right) e^{-\lambda t}\) (b) \(A=\left(\frac{m N_{A} \lambda}{M}\right) e^{-\lambda t}\) (c) \(A=\left(\frac{m N_{A} \lambda}{t}\right) e^{-\lambda t}\) (d) \(A-\left(\frac{M m \lambda}{N_{A}}\right) e^{-\lambda t}\)

Which of the following nuclei is produced when a \({ }_{92} \mathrm{U}^{238}\) nucleus undergoes a \((d, 2 n)\) reaction followed by a beta decay? (a) \({ }_{93} \mathrm{~Np}^{238}\) (b) \({ }_{94} \mathrm{Pu}^{239}\) (c) \({ }_{94} \mathrm{Pu}^{238}\) (d) \({ }_{92} \mathrm{U}^{238}\)

\(\beta\) -emission must be assocrated with : (a) neutrino emission (2) anti-neutrino emission (c) positron emission (d) proton emission
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