Question

The average translational kinetic energy of a molecule in a gas becomes equal to 0.49 eV at a temperature about (Boltzmann constant = 1.38 x 10^-23 JK^-1) ____________.

पर्याय

• 4370° C

• 3514° C

• 5333° C

• 5060° C

Answer 1

The average translational kinetic energy of a molecule in a gas becomes equal to 0.49 eV at a temperature about 3514° C.

Explanation:

`"K.E" _"avg" = 3/2 "k"_"B""T"`

`therefore "T" = 2/3 xx (0.49 xx 1.6 xx 10^-19)/(1.38 xx 10^-23)`

` = 3787.44 "K"`

`= 3514° "C"`