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


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