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

`lambda = h/(mv)`
`1/2 mv^2 = KE`
`mv^2 = 2KE`
`m^2v^2 = 2m KE`
`:. mv = sqrt(2 m KE)`
`:. lambda = h/(sqrt(2m.KE))`
Now `K.E. prop T`
`lambda prop h/(sqrt(2 mT))`
`lambda_(He) prop h/(sqrt(2m_(He)T_(He)))`
`lambda_(Ne) prop h/(sqrt(2m_(Ne)T_(Ne)))`
`(lambda_(He))/(lambda_(Ne)) prop sqrt((m_(Ne)T_(Ne))/(m_(He)T_(He)))`
`T_H = 273 - 73 = 200 K`,
`T_(Ne) = 727 + 273 = 1000`
`(lambda_(He))/(lambda_(Ne)) = sqrt((20 xx 1000)/(4 xx 200))`
`(lambda_(He))/(lambda_(Ne)) = 5`
`:. lambda_(He) = 5 lambda_(Ne)`.