At what temperature the root mean square velocity is equal to escape velocity from the surface of earth for hydrogen and for oxygen ? Given radius of earth ` = 6.4 xx 10^(6) m , g = 9.8 ms^(-2)`. Boltzmann constant `= 1.38 xx 10^(-23) J "molecule"^(-1)`.

Answer

`1.01 xx 10^94)K`, `1.61 xx 10^(5)K`

Explanation

Escape velocity, `upsilon_(e) = sqrt(2gR)`,
given , `C =upsilon_(e) = sqrt(2gR)`
From kinetic interpretation of temperature,
`1/2 mC^(2) = 3/2 kT`
or `1/2 m(2gR) = 3/2 kT`
or `T=(2mgR)/(3k)`
for hydrogen, `T_(H) = (2m_(h)gR)/(3k)`
`=(2 xx ((2)/(6.02 xx 10^(23))xx10^(-3))xx9.8 xx 6.4 xx 10^(6))/(3 xx 1.38 xx 10^(-23))`
`=1.01 xx 10^94)K`
For oxygen ,
`T_(O)=(2m_(0)gR)/(3k)`
`=(2 xx (32/(6.02 xx 10^(23))xx10^(-3))xx9.8 xx 6.4 xx 10^(6))/(3 xx 1.38 xx 10^(-23))`
`=1.61 xx 10^(5)K`