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A parallel stream of hydrogen atoms with velocity v==600 m//s falls normally on a diaphragm with a narrow slit behind which a screen is placed at a distance l= 1.0m. Using the uncertainty principle, evaluate the width of the slit delta at which the width of its image on the screen is minimum. |
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Answer» Solution :Suppose the width of the slit (its extension along the `y`-axis) is `delta`. Then each electron has an UNCERTAINTY `Delta y~delta`. This translates to an uncertainty `DeltaP_(y)~ħ//delta`. We must therefore have `p_(y)underset(~)gtħ//delta`. For the image, brodering has two sources. We write `Delta(delta)=delta+Delta'(delta)` where `Delta'` is the width caused by the spreading of electrons due to their TRANSVERSE momentum. We have `Delta' = v_(y) (l)/(v_(x)) ~= P_(y)(l)/(p) = (lħ)/(m v delta)` Thus `Delta(delta)=delta+(l ħ)/(MV delta)` For large `delta, Delta(delta)~ delta` and quantum effect is UNIMPORTANT. For small `delta`, quantum effects are large. But `Delta (delta)` is minimum when `delta=sqrt((l ħ)/(mv))` as we see by completing the square. Substitution gives `delta= 1.025xx10^(-5)~~ 0.01mm` |
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