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3 years ago

9

Lily took 57 seconds to walk from classroom to library. If the distance between the classroom and library was 38 m, at what aver

age speed did Lily walk in m/min?
___ m/min

1 answer:

slamgirl [31]3 years ago

4 0

1.5 sorry if I’m wrong!

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An atom of helium has a radius of 31. pm and the average orbital speed of the electrons in it is about 4.4x 10 m/s Calculate the

otez555 [7]

The question is incomplete, here is the complete question:

An atom of helium has a radius of 31. pm and the average orbital speed of the electrons in it is about $4.4\times 10^6m/s$ . Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of helium. Write your answer as a percentage of the average speed, and round it to 2 significant digits

<u>Answer:</u> The percentage of average speed is 41 %

<u>Explanation:</u>

We are given:

Radius of helium atom = 31 pm = $31\times 10^{-12}m$ (Conversion factor: $1m=10^{12}pm$ )

So, diameter of helium atom = $(2\times r)=(2\times 31\times 10^{-12})=64\times 10^{-12}m$

The diameter of the atom will be equal to the uncertainty in position.

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p=\frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = d = $64\times 10^{-12}m$

$\Delta p$ = uncertainty in momentum = $m\Delta v$

m = mass of electron = $9.1095\times 10^{-31}kg$

h = Planck's constant = $6.627\times 10^{-34}kgm^2/s^2$

Putting values in above equation, we get:

$64\times 10^{-12}m\times 9.1095\times 10^{-31}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 64\times 10^{-12}m\times 9.1095\times 10^{-31}kg}=1.81\times 10^6m/s$

To calculate the percentage of average speed, we use the equation:

$\text{Percentage of the average speed}=\frac{\text{Uncertainty in velocity}}{\text{Average orbital speed}}\times 100$

We are given:

Average orbital speed = $4.4\times 10^6m/s$

Putting values in above equation, we get:

$\text{Percentage of the average speed}=\frac{1.81\times 10^6m/s}{4.4\times 10^6m/s}\times 100\\\\\text{Percentage of the average speed}=41.\%$

Hence, the percentage of average speed is 41 %

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