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In Exercise 45, the case is made that the position uncertainty for a typical macroscopic object is generally so much smaller than its actual physical dimensions that applying the uncertainty principle would be absurd. Here we gain same idea of how small an object would have鈾 to be before quantum mechanics might rear its head. The density of aluminum is 2.7103kg/m3, is typical of solids and liquids around us. Suppose we could narrow down the velocity of an aluminum sphere to within an uncertainty of1mper decade. How small would it have to be for its position uncertainty to be at least as large as110%of its radius?

Short Answer

Expert verified

The radius of the microscopic object to obtain the required uncertainty in position is r105m.

Step by step solution

01

Concept of Heisenberg's Uncertainty Principle. 

The expression for density is given by,

D=m43蟺搁3m=43蟺搁3D

Here, mis mass, Dis the density and Ris the radius

The expression for Heisenberg鈥檚 uncertainty principle in terms of mass and velocity is given by,

尘螖惫x螖虫h2(43蟺搁3D)螖惫x螖虫h28蟺顿搁3螖惫x螖虫3h螖虫3h8蟺顿搁3螖惫x

The expression for uncertainty in position as a fraction of radius is given by,

螖虫Rk(38eDR3螖惫x)Rk3h8蟺顿搁4螖惫xkR3h8蟺kD螖惫x

Here, 螖惫xthe change in velocity, Dis the density, h is Planck鈥檚 constant, 螖虫

is the uncertainty in position and Ris the radius.

02

Use Heisenberg’s Uncertainty Principle for calculation. 

Consider volume for the object isV=43r5and the uncertainty in velocity, consequently the momentum, is known.

Substitute given values we can get an estimate of the value ofr(notex=0.10%of r).

螖虫螖辫2螖虫尘螖惫2

Here,xis the uncertainty in position, 螖辫is uncertainty in momentum, m is the mass and vis velocity change and is Planck鈥檚 constant.

x=0.001r,鈥夆赌m=Vm=43r3

Now substitute the value in formula

螖虫尘螖惫20.001r43蟺谤3螖惫2

Solve further as:

r20.00143螖惫4r=1.0541034kgm2s20.0012.7103kg/m343106m10365246060s4r=1.46910204mr105m

Thus, the radius of the microscopic object to obtain the required uncertainty in position isr105m .

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