The atomic masses of He and Ne are 4 and 20amu , respectively. The...

DeBroglie wave length is given as
λ = \dfrac{\text{h}}{\sqrt{2 \text{m} \left(\text{K.E}\right)}} = \dfrac{\text{h}}{\sqrt{2 \text{m} \dfrac{3}{2} {\text{k}}_{\text{b}} · \text{T}}} = \dfrac{\text{h}}{\sqrt{{\text{3 k}}_{\text{b}} \text{m T}}}
${\text{k}}_{\text{b}}$ is Boltzmann's constant
{&lambda;}_{\text{He}} = \dfrac{\text{h}}{\sqrt{{\text{3 k}}_{\text{b}} \left(2 0 0\right) 4 × \text{1.66} × 1 {0}^{- 2 7}}} }$\begin{array}{ll}\text{T} = \text{200 K} \\ {\text{m}}_{\text{He}} = 4 × \text{1.66} × 1 {0}^{- 2 7} \text{kgs}\end{array}$<br />${&lambda;}_{\text{Ne}} = \dfrac{\text{h}}{\sqrt{{\text{3 k}}_{\text{b}} \left(1 0 0 0\right) 2 0 × \text{1.66} × 1 {0}^{- 2 7}}} }\begin{array}{ll}\text{T} = \text{1000 K} \\ {\text{m}}_{\text{Ne}} = 2 0 × \text{1.66} × 1 {0}^{- 2 7} \text{kgs}\end{array}$$$<br />$\dfrac{{λ}_{\text{He}}}{{λ}_{\text{Ne}}} = \sqrt{\dfrac{1 0 00 × 2 0}{2 0 0 × 4}} = \sqrt{2 5} = 5$