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An electron has a velocity of 3.2 x 10^6 m/s. What is its’ momentum? (b) What is its’ wavelength? (c) What other objects/materials have this space/size? (d) Assuming that we can measure the velocity to an accuracy of 10%. Use the Heisenberg uncertainty principle to calculate the uncertainty in the position.

Respuesta :

Answer:

P = 2.91*10^{-24} kg m/s

[tex]\lambda = 2.73 *10^{-10} m[/tex]

size of atom hat lie in range of 1 to 5 Angstrom

[tex]\Delta x = 0.2272[/tex] Angstrom

Explanation:

A) MOMENTUM

p = mv

where m is mass of electron

so momentum p can be calculated as

p = 9.11*10^{-31} *3.2*10^{6}

P = 2.91*10^{-24} kg m/s

b) wavelength

[tex]\lambda = \frac{h}{mv}[/tex]

where h is plank constant

so[tex] \lambda = \frac{6.626*10^{-34}}{2.91*10^{-24}}[/tex]

[tex]\lambda = 2.73 *10^{-10} m[/tex]

c) size of atom hat lie in range of 1 to 5 Angstrom

d) from the information given in the question we have

[tex]\frac{\Delta v}{v} = 0.1[/tex]

[tex]\Delta v = 0.1 v[/tex]

we know that

[tex]\Delta p *\Delta x = \frac{h}{4\pi}[/tex]

[tex]m \Delta v \Delta x =\frac{h}{4\pi}[/tex]

[tex]\Delta x = \frac{h}{m \Delta v}[/tex]

[tex]\Delta x  = \frac{2.272}{0.1}[/tex]                      [[tex]\Delta v = 0.1 v[/tex]]

[tex]\Delta x = 0.2272[/tex] Angstrom

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