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Question

An electron is allowed to move freely in a closed cubic box of length of side 10 mm. The uncertainity in its velocity will be:

A
3.35×103 ms1
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B
5.8×104 ms1
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C
4×105 ms1
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D
4×106 ms1
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Solution

The correct option is A 3.35×103 ms1
We know that from Heisenberg's uncertainity principle,
Δx.Δph4π
where Δx:uncertainity in position
Δp:uncertainity in momentum
h:planck's constant
Side of cube =10 mm=10×103 m
Maximum uncertainity in position = Length of body diagonal of cube. Because it is the maximum possible length in the cube.
Δx=103 mm=103×103 m
Here, p=mv
Δp=Δ(mv)
Δp=mΔv
Δvh4πmΔx
6.626×10344×3.14×(9.1×1031)×(103×103)
Δv3.34×103 m/s

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