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If the uncertainty in the position of a ...

If the uncertainty in the position of a particle is equal to its de-Broglie wavelength, the minimum uncertainty in its velocity should be

A

`1/(4 pi)`

B

`v/(4pi)`

C

`v/(4pi m)`

D

`(mv)/(4pi)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to follow these steps: ### Step 1: Understand the given information We are given that the uncertainty in the position of a particle (\( \Delta x \)) is equal to its de Broglie wavelength (\( \lambda \)). ### Step 2: Write the de Broglie wavelength formula The de Broglie wavelength (\( \lambda \)) is given by the formula: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) can also be expressed in terms of mass (\( m \)) and velocity (\( v \)): \[ p = mv \] Thus, we can rewrite the de Broglie wavelength as: \[ \lambda = \frac{h}{mv} \] ### Step 3: Use the uncertainty principle According to the Heisenberg uncertainty principle, the product of the uncertainties in position and momentum is given by: \[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \] Here, \( \Delta p \) can be expressed as: \[ \Delta p = m \cdot \Delta v \] where \( \Delta v \) is the uncertainty in velocity. ### Step 4: Substitute \( \Delta x \) and \( \Delta p \) in the uncertainty principle Substituting \( \Delta x = \lambda \) and \( \Delta p = m \cdot \Delta v \) into the uncertainty principle gives: \[ \lambda \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \] ### Step 5: Substitute for \( \lambda \) Now, substituting \( \lambda = \frac{h}{mv} \) into the equation: \[ \frac{h}{mv} \cdot (m \cdot \Delta v) \geq \frac{h}{4\pi} \] ### Step 6: Simplify the equation This simplifies to: \[ \frac{h \Delta v}{v} \geq \frac{h}{4\pi} \] Now, we can cancel \( h \) from both sides (assuming \( h \neq 0 \)): \[ \frac{\Delta v}{v} \geq \frac{1}{4\pi} \] ### Step 7: Solve for \( \Delta v \) Multiplying both sides by \( v \) gives: \[ \Delta v \geq \frac{v}{4\pi} \] ### Final Answer Thus, the minimum uncertainty in the velocity (\( \Delta v \)) is: \[ \Delta v = \frac{v}{4\pi} \] ---
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