Show that if the measurement of the uncertainty in the location of the particle is equal to its de Broglie wavelength, the minimum uncertainty in the its velocity `(DeltaV)` Is equal to `1//4pi` of its velocity (V).

Explanation

`Delta x =`
`Delta v = ?`
`Deltax . Deltap ge h/(4pi)`
`lambda(m Delta v) ge h/(4 pi)`
`Deltav ge (h)/(4pi (m lambda))`
`Delta v > (h)/(4 pi xx m xx h/(mv)) " "[ :. lambda = h/(mv)]`
`Delta v ge v/(4pi)`
Therefore,
minimum uncertainty in velocity = `v/(4pi)`.