If the velocity of the electron in Bohr's first orbit is `2.19xx10^(6) m s^(-1)`, calculate the de Broglie wavelength associated with it.

Explanation

According to de Broglie’s equation,
`lambda = (h)/(mv)`
Where,
λ = wavelength associated with the electron
h = Planck’s constant
m = mass of electron
v = velocity of electron
Substituting the values in the expression of `lambda` :
`lambda = (6.626 xx 10^(-34) Js)/((9.10939 xx 10^(-31) kg)(2.19 xx 10^(6) ms^(-1)))`
`= 3.32 xx 10^(-10)m = 3.32 xx 10^(-10) m xx (100)/(100)`
`= 332 xx 10^(-12) m`
`lambda = 332 "pm"`
`:.` Wavelength associated with the electron `= 332 "pm"`