Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 106 ms–1, calculate de Broglie wavelength associated with this electron.
Given:
Velocity of electron, v = 1.6 × 106 ms–1
Mass of electron, m = 9.1× 10-31 kg
According to the de Broglie’s Equation:
λ = h/mv
Where,
λ = wavelength of moving particle
m = mass of electron
v= velocity of particle
h= Planck’s constant [6.62× 10-34]
λ = {[6.62× 10-34] / [9.1× 10-31] × [1.6 × 106]}
= {[6.62× 10-34] / [1.456 × 10-24]
= 4.5508× 10-10 m
Therefore, the wavelength is 455 pm