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Question

Why is the following situation impossible? An air rifle is used to shoot 1.00-g particles at a speed of vx = 100 m/s. The rifle’s barrel has a diameter of 2.00 mm. The rifle is mounted on a perfectly rigid support so that it is fired in exactly the same way each time. Because of the uncertainty principle, however, after many firings, the diameter of the spray of pellets on a paper target is 1.00 cm.

Solution

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Knowns & Concept

The mass of the particles is m=1.00g=1.00×103kgm=1.00\, \text{g}=1.00\times 10^{-3}\, \text{kg}, and their speed is v=100msv=100\, \dfrac{\text{m}}{\text{s}}. The diameter of the hole is d=2.00mm=2.00×103md=2.00\, \text{mm}=2.00\times 10^{-3}\, \text{m}. We need to find the distance xx (see the attached picture). From the Heisenberg uncertainty principle\textbf{the Heisenberg uncertainty principle}, we can write:

dΔpy2d\Delta p_y \geq \dfrac{\hbar}{2}

and we choose the case when dΔpyd\Delta p_y has the smallest value 2\dfrac{\hbar}{2}, where =h2π\hbar=\dfrac{h}{2\pi} is the reduced Planck's constant h=6.626×1034Jsh=6.626 \times 10^{-34}\, \text{Js}. Also, from the similarity of the blue and the green triangles at the picture, we can write:

1cmx=Δpypx\dfrac{1\, \text{cm}}{x}=\dfrac{\Delta p_y}{p_x}

where px=mvp_x=mv is the momentum of the particles along xx axis. Now, we will express Δpy\Delta p_y as:

Δpy=px×1cmx\Delta p_y=\dfrac{p_x\times 1\, \text{cm}}{x}

and substitute pyp_y at the Heisenberg principle:

d×px×1cmx=2d\times \dfrac{p_x\times 1\, \text{cm}}{x}= \dfrac{\hbar}{2}

Rearrange:

x=2dpx×1cm=4πdpx×1cmhx=\dfrac{2dp_x \times 1\, \text{cm}}{\hbar}=\dfrac{4\pi dp_x \times 1\, \text{cm}}{h}

Substitute numerical values:

x=4×3.14×2.00×103m×1.00×103kg×100ms×1×102m6.626×1034Jsx=\dfrac{4\times 3.14\times 2.00\times 10^{-3}\, \text{m}\times 1.00\times 10^{-3}\, \text{kg} \times 100\, \dfrac{\text{m}}{\text{s}}\times 1\times 10^{-2}\, \text{m}}{6.626 \times 10^{-34}\, \text{Js}}

x=3.79×1028m\boxed{x=3.79\times 10^{28}\, \text{m}}

Clearly, the required distance is larger than the diameter of the visible Universe D=2×1026mD=2\times 10^{26}\, \text{m}.

xD=3.79×10282×1026=190\dfrac{x}{D}=\dfrac{3.79\times 10^{28}}{2\times 10^{26}}=190

The distance x is 190 times larger than the observable Universe!

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