\(E _{ O _{2}}=0.048 \; eV\)

We know that,

Translational kinetic energy,

\(E =\frac{3}{2} KT\)

where, \(k =\) Boltzmann constant

\(T =\) Temperature

\(\therefore E \propto T\)

Since, the temperature is same for both oxygen and nitrogen therefore, \(E _{ O _{2}}= E _{ N _{2}}\)

\(\therefore E _{ N _{2}}=0.048 \; eV\)