Here work function, ϕ₀ = 0.5 eV
According to Einstein's photoelectric equation

Maximum kinetic = Incident – Work function
the energy of the emitted photon
electrons energy

∴  $K_{ma{x}_1} = 1. \mathrm{eV} – 0.5 \mathrm{eV} = 0.5 \mathrm{eV}$             ...(i)

and $K_{ma{x}_2} = 2.5 \mathrm{eV} – 0.5 \mathrm{eV} = 2 \mathrm{eV}$             ...(ii)

Divide (i) by (ii), we get

$\frac{{\text{K}}_{{\text{max}}_1}}{{\text{K}}_{{\text{max}}_2}}=\frac{0·5\text{eV}}{2\text{eV}}=\frac{1}{4}$

$\frac{\frac{1}{2}{\text{mv}}_{{\text{max}}_1}^2}{\frac{1}{2}{\text{mv}}_{{\text{max}}_2}^2}=\frac{1}{4}\text{or}\frac{{\text{v}}_{{\text{max}}_1}}{{\text{v}}_{{\text{max}}_2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$