The average kinetic energy of a monatomic molecule is $$0.414$$ eV at temperature: (Use $$K_B = 1.38 \times 10^{-23} \text{ J mol}^{-1} \text{ K}^{-1}$$)

The average kinetic energy of a monatomic molecule is:

$$KE = \frac{3}{2}k_BT$$

Given $$KE = 0.414$$ eV $$= 0.414 \times 1.6 \times 10^{-19} = 6.624 \times 10^{-20}$$ J.

$$T = \frac{2 \times KE}{3k_B} = \frac{2 \times 6.624 \times 10^{-20}}{3 \times 1.38 \times 10^{-23}} = \frac{13.248 \times 10^{-20}}{4.14 \times 10^{-23}} = \frac{13.248}{0.0414} \approx 3200 \text{ K}$$

The answer is $$3200$$ K, which corresponds to Option (2).