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kifflom [539]
3 years ago
14

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/materia

ls 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.
Physics
1 answer:
Romashka [77]3 years ago
7 0

Answer:

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

\lambda = 2.73 *10^{-10} m

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

\Delta x = 0.2272 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

\lambda = \frac{h}{mv}

where h is plank constant

so\lambda = \frac{6.626*10^{-34}}{2.91*10^{-24}}

\lambda = 2.73 *10^{-10} m

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

d) from the information given in the question we have

\frac{\Delta v}{v} = 0.1

\Delta v = 0.1 v

we know that

\Delta p *\Delta x = \frac{h}{4\pi}

m \Delta v \Delta x =\frac{h}{4\pi}

\Delta x = \frac{h}{m \Delta v}

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

\Delta x = 0.2272 Angstrom

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