JEE Main 2015 · 11 Apr · Shift 1, Chemistry question 33

A visible photon has $\lambda\approx400$-700 nm. A thermal particle has $\lambda=\dfrac h{\sqrt{3mkT}}$: about 6 nm for an electron at 300 K and about 0.15 nm for a neutron. So visible photon > thermal electron > thermal neutron.