Answer 1

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 λ:

`lambda = (6.626 xx 10^(-34) Js)/((9.10939 xx 10^(-31)kg)(2.19xx10^6 ms^(-1))`

`= 3.32 xx 10^(-10)m = 3.32 xx 10^(-10) mxx 100/100`

`= 332xx10^(-12) m`

λ = 332 pm

∴Wavelength associated with the electron = 332 pm