Calculate the De Broglie wavelength of a 1 times 10^9 kg (that\'s 1 microgram) d

Question

Calculate the De Broglie wavelength of a 1 times 10^9 kg (that's 1 microgram) dust particle that has a kinetic energy of 1000 eV. Give your answer in nanometers. 1.17 times l0^-21 m = 1.17 times 10^-12 nm Does this De Broglie wavelength have any physical meaning on the sub nanometer dimensions of the subatomic world? Why? The De Broglie wavelength is MUCH smaller than the size of atoms of atoms Calculate the De Broglie wavelength of an electron (mass = 9.11 times 10^-31 kg) that has a kinetic energy of 1000 eV. Give your answer in nanometers. 3.88 times 10^-11 m = .038 nm Does this De Broglie wavelength have any physical meaning on the sub nanometer dimensions of the subatomic world? Why? YES - The De Broglie wavelength is on the scale of the size of atoms Compare and contrast the particle nature (classical physics) and quantum (wave) nature of particles. Be sure to include any equations that are relevant. What I am looking for is some version of the 'big chart" I put on the board in class. .. In a nutshell: Classical: exact position and velocity can be known. 1 cause = 1 effect, can predict the future Quantum: exact position and velocity can NOT be known, 1 cause = many possible effects, can NOT accurately predict the future. Wave representing the particle: Amplitude (really A^2) proportional to probability of finding particle at a location, wavelength (De Broglie wavelength) = particle fuzziness, sphere of influence, region it is likely to be found in, etc.

Explanation / Answer

5) a1) Debroglie's wavelength is lamda = h/sqrt(2*m*E)

h is planck;s constant = 6.626*10^-34 m^2 kg /sec

m is the mass = 1*10^-9 Kg
E = 1000 eV = 1000*1.6*10^-19

lamda = 6.626*10^-34/sqrt(2*10^-9*1.6*10^-19*1000) = 1.17*10^-21 m = 1.17*10^-12 nm

a2) size of the atom is in the order of 10^-15 m ,but here lamda is in the order of 10^-21 m

so the answer is No

b1) lamda = h/sqrt(2*m*E) = 6.626*10^-34/sqrt(2*9.11*10^-31*1.6*10^-19*1000) = 3.88*10^-11 m = 0.038 nm

b2) since the lamda is in the order od 10^-11 and atom is also comparable size(10^-15)

so the answer is yes

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