(i) λiron ball = \(\frac{h}{mv}\)
= 1 x 10-35m
(ii) λiron ball = \(\frac{h}{mv}\)
= \(\frac{6.626}{662.6}\) x 10-3
= 1 x 105 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