Question

If the velocity of the electron in Bohr’s first orbit is 2.19×106ms–1, calculate the de Broglie wavelength associated with it.

Answer 1

According to de Broglie's equation,
λ = $\frac{h}{mv}$
Where, λ = wavelength associated with the electron
h= Planck's constant
m= mass of the electron
v= velocity of electron
Substituting the values in the expression of λ :
λ = $\frac{6.626 × 10^{-34}Js}{(9.10939 × 10^{-31}kg)(2.19 × 10^6 ms^{-1})}$
= 3.32 × 10-10 m = $\frac{3.32\times 10^{-10}m\times 100}{100}$
= 332 × 10-12 m
λ = 332 pm
∴ The wavelength associated with the electron = 332 pm