Question 2.58 Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
Class XI Structure of Atom Page 68
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wavelength ( λ ) = h/mv
mass of neutron ( m ) = 1.675 × 10^-27 Kg
wavelength ( λ ) = 800 pm = 8 × 10^-10 m
h = 6.626 × 10^-34 Js
now, λ = h/mv
8 × 10^-10 m = 6.626 × 10^-34 Js/1.675 × 10^-27 Kg × v
v = 8 × 10^-10 × 1.675 × 10^-27/8 × 10^-10
= 0.494 × 10³ m/s = 4.94 × 10² m/s
hence, velocity of electron = 4.94 × 10² m/s
mass of neutron ( m ) = 1.675 × 10^-27 Kg
wavelength ( λ ) = 800 pm = 8 × 10^-10 m
h = 6.626 × 10^-34 Js
now, λ = h/mv
8 × 10^-10 m = 6.626 × 10^-34 Js/1.675 × 10^-27 Kg × v
v = 8 × 10^-10 × 1.675 × 10^-27/8 × 10^-10
= 0.494 × 10³ m/s = 4.94 × 10² m/s
hence, velocity of electron = 4.94 × 10² m/s
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