Discussion Forum

 The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73° C is "M"  times that of the de-Broglie wavelength of Ne at 727° C. M is 

Answer:

Explanation:

PLAN     $KE=\frac{1}{2}mv^{2}=\frac{3}{2}RT$

  $\therefore$      $m^{2}v^{2}=2mKE$

     $\therefore$       mv=  $\sqrt{2mKE}$

  $\lambda$   (wavelength)=

$\frac{h}{mv}=-\frac{h}{\sqrt{2mKE}}=\frac{h}{\sqrt{2m(T)}}$

$\lambda$(He  at -73° C=200K)=  $\frac{h}{\sqrt{2\times 4\times200}}$

 $\lambda$ (Ne at 727° C=1000k)

  =    $\frac{h}{\sqrt{2\times 20\times1000}}$

 $\therefore$     $\frac{\lambda_{(He)}}{\lambda_{(Ne)}}=\sqrt{\frac{2\times20\times1000}{2\times4\times200}}=5$

   Thus, M=5