If the Kinetic energy of a sub – atomic particle is increases 8 times, its de – Brogllie wavelength x times the original wavelength. The value of x is

de- Broglie relation is $m\upsilon =h/\lambda$.The kinetic energy becomes

            $T=\frac{1}{2}m{\upsilon }^2=\frac{h^2}{2m{\lambda }^2}$           Hence, $\frac{T_2}{T_1}=\frac{{\lambda }_1^2}{{\lambda }_2^2}\Rightarrow \frac{8T_1}{T_1}=\frac{{\lambda }_1^2}{{\lambda }_2^2}\Rightarrow {\lambda }_2^2=\frac{{\lambda }_1^2}{8}\Rightarrow {\lambda }_2=\frac{{\lambda }_1}{2\sqrt{2}}$