2). 10pts.: a). Calculate the wavelength of an electron with a velocity of 2.99

Question

2). 10pts.: a). Calculate the wavelength of an electron with a velocity of 2.99 x10^6 m/s b). Calculate the wavelength of an neutron with a velocity of 2.99 x10^6 m/s (Hint: use the mass of the neutron and electron on page 1). The uncertainty in the velocity of both the electron and the neutron is +/- 1.00 x 10^5 m/s, roughly 3.3%. c). Calculate the uncertainty in the position of the electron. Ignore relativistic effects. d). Calculate the uncertainty in the position of the neutron. Ignore relativistic effects. e). What implication does this have to our understanding of atomic structure?

Explanation / Answer

2)

a) using debroglie wavelenght equation we get

wavelength lamda = h / mv

so

lamda for electron = 6.634 x 10-34 / 9.1 x 10-31 x 2.99 x 106

lamda for electron = 0.2438 x 10-9 m

so

wavelength of electron = 0.2438 nm

b) for neutron

wavelength = 6.634 x 10-34 / 1.6749 x 10-27 x 2.99 x 106

wavelength = 1.32 x 10-13 m

c)

according to heisenberg principle

delta x * delta p = h/4pi

delta x = ( h/ 4pi ) / delta p

but we know that

delta p = m * delta v

so

delta x = ( h/4pi) / m * delta v

so for electron

delta x = ( 6.634 x 10-34 / 4 x 3.14 ) / 9.1 x 10-31 x 1 x 10^5

delta x = 5.8 x 10-10

so the uncertainity in the position of electron is 5.8 x 10-10 m

d)

for nuetron

delta x = ( 6.634 x 10-34 / 4 x 3.14 ) / ( 1.6749 x 10-27 x x 10^5 )

delta x = 3.15 x 10-13

so the uncertainity in the position of nuetron is 3.15 x 10-13 m

e)

from the above results , the position of electron is more uncertain than that of nuetron

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