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)
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
Comments
Leave a comment