Question

An electron is confined to a box of length 10-8(to the power -8) m. Calculate the minimum uncertainty

in the velocity. (Ans: Δv = 5.8 km/ s)


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Answer 1

Answer:

according to Heisenberg uncertainty principleΔvΔ x = h / 4 π

so by putting values of h Δx and Pi we get Δv = 5.8 kilometre per second

h = Planck's Constant

Answer 2

The minimum uncertainty in the velocity is 5.8 km/s.

Explanation:

Given that,

An electron is confined to a box of length $10^{-8}\ m$

We need to find the minimum uncertainty in the velocity of the particle. The Uncertainty principle is given by :

$\Delta x{\cdot} \Delta p\ge \dfrac{h}{4\pi}$

h is Planck's constant

$\Delta x.\Delta (mv)\ge \dfrac{h}{4\pi}\Delta p\ge \dfrac{h}{4\pi m\Delta x}\\\Delta p\ge \dfrac{6.63\times 10^{-34}}{4\pi\times 9.1\times 10^{-31}\times 10^{-8}}\\\Delta p\ge 5.27\times 10^{-25}\ kg-m/s\\\Delta p\ge 5797.78\ m/s\\\\\Delta p\ge 5.8\ km/s$

So, the minimum uncertainty in the velocity is 5.8 km/s.

Learn more,

Heisenberg's Uncertainty Principle

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