Question

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.

Answer 1

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