1. Use the equipartition theorem

   For an ideal gas in thermal equilibrium, the equipartition theorem states:

   Each independent quadratic degree of freedom contributes an average energy of $\frac{1}{2}k_B T$

2. Interpret “one degree of freedom”

   A single translational degree of freedom corresponds to kinetic energy of the form $\frac{p_x^2}{2m}$ or equivalently $\frac{1}{2}mv_x^2$

   This is a quadratic term, so its average contribution is $\frac{1}{2}k_B T$

3. Therefore

   The average energy along one degree of freedom is $\boxed{\frac{1}{2}k_B T}$

4. Check options

   • A: $\frac{1}{2}k_B T$ ✅
   • B: $\frac{2}{3}k_B T$ ❌
   • C: $\frac{3}{2}k_B T$ ❌ (this is for 3 translational degrees of freedom combined)
   • D: $k_B T$ ❌

Hence, the correct option is A.