(i) λ_{iron ball }= \(\frac{h}{mv}\)

= 1 x 10^{-35}m

(ii) λ_{iron ball }= \(\frac{h}{mv}\)

= \(\frac{6.626}{662.6}\) x 10^{-3}

= 1 x 10^{5 }m

For an electron, the de Broglie wavelength is significant and measurable while for iron ball it is too small to measure, hence it becomes insignificant