EmSAT Achieve Exam > EmSAT Achieve Questions > Which of the following decay processes is all...
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Which of the following decay processes is allowed?
- a)K0 → µ+ + µ-
- b)μ- → e- + γ
- c)n → p + π -
- d)n → π+ + π-
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Which of the following decay processes is allowed?a)K0→ µ++...
→ For the first process K0 → µ+ + µ-
Thus, this is an allowed decay through the weak interaction. In weak interaction strangeness and
Iso spin is not conserved.
Hence the correct answer is option 1.
Thus, this is an allowed decay through the weak interaction. In weak interaction strangeness and
Iso spin is not conserved.
Hence the correct answer is option 1.
Most Upvoted Answer
Which of the following decay processes is allowed?a)K0→ µ++...
Explanation:
K0→ µ++ µ-
- This decay process is allowed because it conserves both charge and lepton number.
- The initial state, K0, is a neutral meson with zero charge and lepton number.
- The final state, µ++ µ-, consists of two muons, which are charged leptons.
- The decay process conserves charge as the total charge of the initial and final states is zero.
- It also conserves lepton number as the total lepton number of the initial and final states is zero.
- Therefore, this decay process is allowed.
Explanation of other options:
- μ-→ e-+ γ: This process violates lepton flavor conservation and is not allowed.
- n→p +π-: This process violates conservation of baryon number and charge, and is not allowed.
- n→π++π-: This process violates conservation of baryon number and is not allowed.
In summary, option A (K0→ µ++ µ-) is the only allowed decay process out of the given options because it conserves charge and lepton number.
K0→ µ++ µ-
- This decay process is allowed because it conserves both charge and lepton number.
- The initial state, K0, is a neutral meson with zero charge and lepton number.
- The final state, µ++ µ-, consists of two muons, which are charged leptons.
- The decay process conserves charge as the total charge of the initial and final states is zero.
- It also conserves lepton number as the total lepton number of the initial and final states is zero.
- Therefore, this decay process is allowed.
Explanation of other options:
- μ-→ e-+ γ: This process violates lepton flavor conservation and is not allowed.
- n→p +π-: This process violates conservation of baryon number and charge, and is not allowed.
- n→π++π-: This process violates conservation of baryon number and is not allowed.
In summary, option A (K0→ µ++ µ-) is the only allowed decay process out of the given options because it conserves charge and lepton number.
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Which of the following decay processes is allowed?a)K0→ µ++ µ-b)μ-→ e-+ γc)n→p +π-d)n→π++π-Correct answer is option 'A'. Can you explain this answer? for EmSAT Achieve 2025 is part of EmSAT Achieve preparation. The Question and answers have been prepared according to the EmSAT Achieve exam syllabus. Information about Which of the following decay processes is allowed?a)K0→ µ++ µ-b)μ-→ e-+ γc)n→p +π-d)n→π++π-Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for EmSAT Achieve 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Which of the following decay processes is allowed?a)K0→ µ++ µ-b)μ-→ e-+ γc)n→p +π-d)n→π++π-Correct answer is option 'A'. Can you explain this answer?.
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