From the de Brogle relationship
`lambda = (h)/(mv)`
For `CH_(4) ,lambda_(ch4) = (h)/(m_(CH4) xx v_(CH4))`…….(i)
For `O_(2),lambda_(O_2) = (h)/(m_(O_2) xx v_(O_2))`……(ii)
Wavelength of `CH_(4)` and `O_(2)` is equal hence
` (h)/(m_(CH_4) xx v_(CH_4)) =(h)/(m_(O_2) xx v_(O_2))`
`rArr ( v_(CH_4))/( v_(O_2))= (m_(O_2))/(mCH_4) = (32)/(16) = 2 `
`:, v_(CH_4) = 2 v_(O_2)`
THe velocity of `CH_(4)` molecule is two times the velocity of `O_(2)`molecule