1.

The atomic masses of 'He' and 'Ne' are 4 and 20 a.m.u 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»


SOLUTION :`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)`


Discussion

No Comment Found