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If the uncertainty in the position of a...

If the uncertainty in the position of a moving electron is equal to its de Broglie wavelength, then its velocity will be completely uncertain. Explain.

Text Solution

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According to available information , `Deltax=lambda`
But ` lambda =(h)/(mv) =(h)/(p) :. P=(h)/(lambda)=(h)/(Deltax) or Delta x =(h)/(p)`
According to uncertainty principle, `DeltaxDeltap ge (h)/(4pi)`.
`:. " " (h)/(p) Deltap ge (h)/(4 pi) or (Deltap)/(p) ge (1)/(4 pi)` (constant)
This means that `Deltap ge p or m DeltaV ge mV or DeltaV ge V`.
This means that the velocity of electron will be completely uncertain.
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Knowledge Check

  • 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/(4pi)`
    B
    `v/(4pi)`
    C
    `v/(4pi m)`
    D
    `(mv)/(4pi)`

  • 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)`

  • The uncertainity in the position of a particle is equal to the de-Broglie wavelength. The uncertainity in its momentum will be

    A
    `h/lambda`
    B
    `(2h)/(3lambda)`
    C
    `lambda/h`
    D
    `(3lambda)/(2h)`