Calculate the de Broglie wavelength (in pm) of a hydrogen atom traveling 435 m/s.?

Question

Do I set this up as E= h/m(v)??? If so how do I go from g of Hydrogen to pm?

Mr. Toni Zulauf Sr. · Accepted Answer

You find the wavelength λ from the de Broglie relation …λ = h / mv… where ...h is Planck’s constant: 6.63 x 10^-34 J.s (or kg.m^2 / s)m is the mass of the hydrogen atom in kg: 1.67 × 10^-27 kgv is the speed in m/s: 435 m/s Because the speed is so slow - a negligible fraction of the speed of light - you don’t need to use the relativistic version of de Broglie’s equation.… so …λ = (6.63 x 10^-34 kg.m^2 / s) / (1.67 × 10^-27 kg)(435 m/s)= 0.00913 x 10^-7 m= 9.13 x 10^-10 m= 913 pmCheck to see if my figures are correct. I have no idea what the actual answer should be: but this seems to be reasonable for a slow-moving H-atom.

Mr. Toni Zulauf Sr. · Answer

de Broglie Eqn for our purposesWavelength λ = h/(m×v)where h is Plank’s constant = 6.626 × 10^-34 J s; m is mass and v is velocity. You must have the mass in kg (SI units) Mass of H atom = 1.008/6.0022 × 10^23 = 1.674 × 10^-24 g = 1.674 × 10^-27kg(6.0022 × 10^23 is the Avogadro const) should we be using the mass of a proton (ignore H-2)? Won't effect answer much.λ = (6.626 × 10^-34)/(1.674 × 10^-27×435) = 9.1 × 10^-10 m s^-1= 9.1 nm need about 0.1 nm (1 Å) to get diffraction by atomic lattices (as in neutron diffraction)

Sigrid O'Kon

Calculate the de Broglie wavelength (in pm) of a hydrogen atom traveling 435 m/s.?

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