Question

An electron is closed in a box of 5 angstrom width. Find out the minimum energy of the electron.​

Answer 1

Answer:

According to Heisenberg's uncertainity principle:

Δx.Δp=4πh

Δx.Δ(mv)=4πh

Δv=4πmΔxh

h=6.626×10−34Js=Planck's constant

Given m=mass of electron=9.1×10−31 kg and Δx=1000A˚=10−7m

upon substitution we get:

Δv=4×3.14×9.1×10−31×10−76.626×10−34

Δv=0.0579×104=5.79×102m/s

Explanation:

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