The ratio of the energies of a moving particle and a photon is 1/100. Their velocities are in the ratio 1/10, then the ratio of their de-Broglie wavelengths is

${\mathrm{E}}_{\text{particle }}=\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\mathrm{mv}\times \mathrm{v}$

$=\frac{1}{2}\mathrm{mv}\times \frac{\mathrm{h}}{\mathrm{m}\times {\mathrm{\lambda }}_{\text{particle }}}=\frac{{\displaystyle \mathrm{hv}}}{{\displaystyle 2{\mathrm{\lambda }}_{\text{particle }}}}$

${\mathrm{E}}_{\text{photon }}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\text{photon }}}$

$\frac{{\mathrm{E}}_{\text{particle }}}{{\mathrm{E}}_{\text{photon }}}=\frac{\frac{\text{ hv }}{2{\mathrm{\lambda }}_{\text{particle }}}}{\frac{\text{ hc }}{{\mathrm{\lambda }}_{\text{photon }}}}$

$\Rightarrow \frac{{\mathrm{E}}_{\mathrm{p}}}{{\mathrm{E}}_{\mathrm{ph}}}=\frac{1}{2}\frac{\mathrm{v}}{\mathrm{c}}\times \frac{{\mathrm{\lambda }}_{\text{ph }}}{{\mathrm{\lambda }}_{\mathrm{p}}}$

$\Rightarrow \frac{1}{100}=\frac{1}{2}\times \frac{1}{10}\times \frac{{\mathrm{\lambda }}_{\mathrm{ph}}}{{\mathrm{\lambda }}_{\mathrm{p}}}$

$\therefore {\mathrm{\lambda }}_{\mathrm{p}}:{\mathrm{\lambda }}_{\mathrm{ph}}=5:1$