For the electron:
Mass, \(m=9.11\times 10^{-31}\;\mathrm{kg}\)
speed \(v=5.4\times 10^6\;\mathrm{m/s}\)
Then momentum,
\(p=mv\)
\(=9.11\times 10^{-31}\times 5.4\times 10^6\)
\(=4.92\times 10^{-24}\;\mathrm{kgm/s}\)
According to the de Broglie equation,
\(\lambda =\frac h p\)
Planck’s constant, \(h=6.62\times 10^{-34}\mathrm{Js}\)
Substituting both the values for calculating de Broglie’s wavelength,
\(\lambda =\frac{\left(6.62\times 10^{-34}\right)}{4.92\times 10^{-24}}\)
\(=13.5\times 10^{-11}\;\mathrm{m}\)