To solve this problem, we will use the de Broglie wavelength formula, which is given by

$$\lambda = \frac{h}{mv}$$

where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, $m$ is the mass of the particle, and $v$ is the velocity of the particle.

However, it's more relevant to use the form of the de Broglie equation that involves temperature, given that kinetic energy ($KE$) at a particular temperature is linked to the velocity of the gas particles. The kinetic energy for a gas particle can be calculated using the equation:

$$KE = \frac{3}{2}k_{B}T$$

where $k_B$ is the Boltzmann constant and $T$ is the temperature in Kelvin. The velocity $v$ of a gas particle can be related to its kinetic energy by the equation:

$$KE = \frac{1}{2}mv^2$$

Solving for $v$ gives:

$$v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{3k_BT}{m}}$$

Substituting $v$ in the de Broglie wavelength formula gives:

$$\lambda = \frac{h}{m\sqrt{\frac{3k_BT}{m}}} = \frac{h}{\sqrt{3mk_BT}}$$

Given that we are comparing helium (He) and neon (Ne) gases, the relative de Broglie wavelengths can be related by the mass of the particles and their temperatures. For helium and neon, respectively, this becomes:

$$\lambda_{He} = \frac{h}{\sqrt{3m_{He}k_BT_{He}}}$$

$$\lambda_{Ne} = \frac{h}{\sqrt{3m_{Ne}k_BT_{Ne}}}$$

Given the temperatures are –73°C for He and 727°C for Ne, we first convert these temperatures to Kelvin:

$$T_{He} = 200K \quad (\text{-73°C + 273})$$

$$T_{Ne} = 1000K \quad (\text{727°C + 273})$$

The ratio of the de Broglie wavelength of He to Ne, $\frac{\lambda_{He}}{\lambda_{Ne}}$, would thus be:

$$\frac{\lambda_{He}}{\lambda_{Ne}} = \frac{\frac{h}{\sqrt{3m_{He}k_BT_{He}}}}{\frac{h}{\sqrt{3m_{Ne}k_BT_{Ne}}}} = \sqrt{\frac{m_{Ne}T_{Ne}}{m_{He}T_{He}}}$$

Substituting the masses ($m_{He}=4$ a.m.u. and $m_{Ne}=20$ a.m.u., with $1$ a.m.u. approximately equal to $1.660539 \times 10^{-27}$ kg, though this conversion is not strictly necessary for a ratio where units cancel out) and the temperatures:

$$\frac{\lambda_{He}}{\lambda_{Ne}} = \sqrt{\frac{20 \times 1000}{4 \times 200}} = \sqrt{\frac{20000}{800}} = \sqrt{25} = 5$$

Therefore, the value of $M$, which is the multiplier for the de Broglie wavelength of He gas at –73°C compared to that of Ne at 727°C, is 5.