The de Broglie wavelength is given by:
$$ \lambda = \frac{h}{\sqrt{2 m k T}} $$
where $h$ is Planck’s constant, $m$ is the mass of the particle, $k$ is Boltzmann’s constant, and $T$ is the temperature in Kelvin. The ratio of wavelengths depends on the square root of the inverse of the temperature ratio. Thus, the ratio of wavelengths at 127°C and 352°C is 5:3.