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2 years ago

6

German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity (

Δ v ) Δ x ≥ h 4 π m Δ v where h is Planck's constant and m is the mass of the object. The mass of an electron is 9.11 × 10 − 31 kg. What is the uncertainty in the position of an electron moving at 2.00 × 10 6 m/s with an uncertainty of Δ v = 0.01 × 10 6 m/s ?

1 answer:

timama [110]2 years ago

8 0

According to the information given, the Heisenberg uncertainty principle would be given by the relationship

$\Delta x \Delta v \geq \frac{h}{4\pi m}$

Here,

h = Planck's constant

$\Delta v$ = Uncertainty in velocity of object

$\Delta x$ = Uncertainty in position of object

m = Mass of object

Rearranging to find the position

$\Delta x \geq \frac{h}{4\pi m\Delta v}$

Replacing with our values we have,

$\Delta x \geq \frac{6.625*10^{-34}m^2\cdot kg/s}{4\pi (9.1*10^{-31}kg)(0.01*10^6m/s)}$

$\Delta x \geq 5.79*10^{-9}m$

Therefore the uncertainty in position of electron is $5.79*10^{-9}m$

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Answer:

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d) $a=24.07mm/s^{2}$

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b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

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$mg>mg-ma_{c}$

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c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$

and let's calculate the normal force:

$N=m(g-a_{c})$

$N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})$

N=586.58N

so now we can calculate the percentage change:

$\%_{change} = \frac{mg-N}{mg}x100\%$

so we get:

$\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%$

$\%_{change}=0.343\%$

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

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r=4,551.91 km

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$a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m$

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