To find the temperature at which the average kinetic energy of a monatomic molecule is $$0.414 \, \text{eV}$$, we use the equation for the average kinetic energy of a molecule in terms of temperature:

$$K_{\text{avg}} = \frac{3}{2} k_B T$$

Where:

• $$K_{\text{avg}}$$ is the average kinetic energy
• $$k_B$$ is the Boltzmann constant, given as $$1.38 \times 10^{-23} \, \text{J/K}$$
• $$T$$ is the temperature in Kelvin

First, we need to convert the average kinetic energy from eV to Joules since the Boltzmann constant is in Joules. The conversion factor is $$1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$$, so:

$$K_{\text{avg}} = 0.414 \, \text{eV} = 0.414 \times 1.6 \times 10^{-19} \, \text{J}$$

Substituting $$K_{\text{avg}}$$ and $$k_B$$ into the equation:

$$\frac{3}{2} k_B T = 0.414 \times 1.6 \times 10^{-19} \, \text{J}$$

Solving for $$T$$:

$$T = \frac{0.414 \times 1.6 \times 10^{-19} \, \text{J}}{\frac{3}{2} \times 1.38 \times 10^{-23} \, \text{J/K}}$$

After performing the calculations:

$$T \approx 3200 \, \text{K}$$

Therefore, the temperature at which the average kinetic energy of a monatomic molecule is $$0.414 \, \text{eV}$$ is approximately 3200 K.