Werner Heisenberg considered the limits of how precisely we can measure the properties of an electron or other microscopic particle. He determined that there is a fundamental limit to how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also true. This is summed up in what we now call the Heisenberg uncertainty principle. The equation is Δx⋅Δ(mv)≥$$h4$$$$π$$The uncertainty in the position or in the momentum of a macroscopic object like a baseball is too small to observe. However, the mass of microscopic object such as an electron is small enough for the uncertainty to be relatively large and significant. If the uncertainties in position and momentum are equal, the uncertainty in the velocity isIf the uncertainty in velocity and position is same, then the uncertainty in momentum will beWhat would be the minimum uncertainty in $$de-Broglie$$ wavelength of a moving electron accelerated by potential difference of $$6$$ volt and whose uncertainty in position is $$722nm$$?

Question

Werner Heisenberg considered the limits of how precisely we can measure the properties of an electron or other microscopic particle. He determined that there is a fundamental limit to how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also true. This is summed up in what we now call the Heisenberg uncertainty principle.  The equation is Δx⋅Δ(mv)≥$$h4$$$$π$$The uncertainty in the position or in the momentum of a macroscopic object like a baseball is too small to observe. However, the mass of microscopic object such as an electron is small enough for the uncertainty to be relatively large and significant. If the uncertainties in position and momentum are equal, the uncertainty in the velocity isIf the uncertainty in velocity and position is same, then the uncertainty in momentum will beWhat would be the minimum uncertainty in $$de-Broglie$$ wavelength of a moving electron accelerated by potential difference of $$6$$ volt and whose uncertainty in position is $$722nm$$?

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Δx Δ$$p=h4$$π⇒Δ$$p2=h4$$π⇒$$m2$$Δ$$v2=h4$$π⇒Δ$$v=12mh$$πΔx $$=h4$$$$π$$m;             Δx  Δ$$p=h4$$$$π$$$$h4$$$$π$$mΔ$$p=h4$$π,       Δ$$p=mh4$$πλD.B.$$=1506Ao=5Aoand$$ Δx⋅hλ$$2$$×Δλ≥$$h4$$π⇒$$1$$π×$$10$$−$$9$$λ$$2$$×Δλ>$$14$$π ⇒Δλ≥$$2.54$$×$$10$$−$$10$$Δλ≥$$0.625Ao$$

Quantum mechanical model of the atom

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If the uncertainties in position and momentum are equal, the uncertainty in the velocity is

$\begin{array}{l} Δx Δp = \frac{h}{4 π} \Rightarrow {Δp}^{2} = \frac{h}{4 π} \\ \Rightarrow m^{2} {Δv}^{2} = \frac{h}{4 π} \Rightarrow Δv = \frac{1}{2 m} \sqrt{\frac{h}{π}} \end{array}$

If the uncertainty in velocity and position is same, then the uncertainty in momentum will be

$\begin{array}{l} Δx = \sqrt{\frac{h}{4 πm}}; Δx Δp = \frac{h}{4 π} \\ \sqrt{\frac{h}{4 πm}} Δp = \frac{h}{4 π}, Δp = \sqrt{\frac{mh}{4 π}} \end{array}$

What would be the minimum uncertainty in de-Broglie wavelength of a moving electron accelerated by potential difference of 6 volt and whose uncertainty in position is $\frac{7}{22}$ nm?

(p=h/λ or △p=(h/λ² ).△λ)

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Quantum mechanical model of the atom

Unlock the full solution & master the concept.

If the uncertainties in position and momentum are equal, the uncertainty in the velocity is

Δx Δp=h4π⇒Δp2=h4π⇒m2Δv2=h4π⇒Δv=12mhπ

If the uncertainty in velocity and position is same, then the uncertainty in momentum will be

Δx =h4πm; Δx Δp=h4πh4πmΔp=h4π, Δp=mh4π

What would be the minimum uncertainty in de-Broglie wavelength of a moving electron accelerated by potential difference of 6 volt and whose uncertainty in position is 722nm?

λD.B.=1506Ao=5Aoand Δx⋅hλ2×Δλ≥h4π⇒1π×10−9λ2×Δλ>14π ⇒Δλ≥2.54×10−10Δλ≥0.625Ao