This answer is restricted. Please login to view the answer of this question.

Login NowHere is given,

Uncertainty in the position of a particle say ∆x = de-Broglie wavelength of the particle (λ)

Uncertainty in the velocity (∆v) = ?

We know that, from the uncertainty principle,

\(∆x ∆p \geq \frac{h}{2x}\)

or, ∆x ∆p = \(\frac{h}{2 \pi ∆v}\)

or, ∆x ∆p = \(\frac{h}{2 \pi m ∆v}\) ——– (1)

Again, From the de-Broglie hypothesis,

\(λ = \frac{h}{mv}\) ——– (2)

According to the question, from the equation (1) and (2)., we get

\(\frac{h}{2\pi m ∆v} = \frac{h}{mv}\)

\(∆v = \frac{v_{wave}}{2\pi}\)

Hence, Uncertainty in velocity is equal to the \(\frac{1}{2\pi}\) times velocity of the de-Broglie wave.

If you found any type of error on the answer then please mention on the comment or report an answer or submit your new answer.

Click here to submit your answer.

HAMROCSIT.COM

## Discussion