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

5

Explanation

`lamda = (h)/(mv)`
`:. (lamda_(He))/(lamda_(Ne)) = (m_(He) xx v_(He))/(m_(He) xx v_(He)) = (M_(Ne) xx v_(Ne))/(M_(He) xx v_(He))`
As `v = sqrt((3RT)/(M)) , (v_(Ne))/(v_(He)) = sqrt((T_(Ne))/(M_(Ne)) xx (M_(He))/(T_(He)))`
`:. (lamda_(He))/(lamda_(Ne)) = (M_(Ne))/(M_(He)) xx sqrt((T_(Ne))/(T_(He)) xx (M_(He))/(M_(Ne)))`
`= sqrt((M_(Ne) T_(Ne))/(M_(He) T_(He))) = sqrt((20 xx 1000)/(4xx 200)) = 5`
`:. lamda_(He) = 5 lamda_(Ne)`