If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1, calculate the de Broglie wavelength associated with it.
Given:
Velocity of electron, v = 2.19 × 106 ms–1
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] × [2.19 × 106]}
= {[6.62× 10-34] / [1.9929 × 10-24]
= 3.3248× 10-10 m
Therefore, the wavelength is 332 pm