Answer to Question #212209 in Chemistry for Bittia

Question #212209

A beam of helium atom moves with a velocity of 4.0*103ms-1. Find the wavelength of the particle constituting the beam. (h=6.626*10-34Js, He=4u)


1
Expert's answer
2021-07-01T05:30:38-0400

Solution:

Gram atomic mass of He = 4 g

Mass of 6.022×1023 atoms of He = 4g = 4×10-3 kg

Mass of 1 atom of He (m) = (4×10-3 kg) / (6.022×1023) = 6.64×10−27 kg


de Broglie equation can be used: λ = h / (mv)

λ = unknown

h = 6.626×10−34 kg m2 s−1

m = 6.64×10−27 kg

v = 4.0×103 m s−1


Thus, according to de Broglie equation:

λ = (6.626×10−34 kg m2 s−1) / (6.64×10−27 kg × 4.0×103 m s−1) = 2.495×10−11 m

λ = 2.495×10−11 m


Answer: The wavelength (λ) of the particle constituting the beam is 2.495×10−11 m

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