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If uncertainty in position and momentum are equal, then uncertainty in velocity is :
Options
(a) 1/2m x √(h/π)
(b) √(h/2π)
(c) 1/m x √(h/π)
(d) √(h/π)
Correct Answer:
1/2m x √(h/π)
Explanation:
We known Δp.Δx ≥ h / 4π
or m.Δv.Δx = h / 4π [ .·. Δp = mΔV]
since Δp = Δx (given)
.·. Δp.Δp = h / 4π or mΔv = h / 4π
or (Δv)² = h / 4πm²
or Δv = √ h / 4πm² = 1 / 2 m √ h / π..
Related Questions: - CCl₄ and freons
- What is the nature of the bond between B and O in (C₂H₅)₂ OBH₃?
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Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Structure of Atom
(90)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- CCl₄ and freons
- What is the nature of the bond between B and O in (C₂H₅)₂ OBH₃?
- The weakest base among the following is
- Haemoglobin is a complex of
- Which of the following compounds is known as white vitriol
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Structure of Atom (90)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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We Know that, Δx.Δp ≥ h/4π,
∵ x = p
we can find answer with using two formulae,
first- Δx.Δp ≥ h/4π,
second- Δx.Δv ≥ h/4mπ.
Here, Δx = Δp = y
According to Heisenberg’s uncertainty principle
Δx*Δp = h/4π
y² = h/4π
y = √h/√4π
y = 1/2 √h/√π
So p(Momentum) = 1/2 √h/√π = mΔv
1/2 √h/√π = mΔv
1/2m √h/√π = Δv
So ans is….. Δv = 1/2m √h/√π