The average kinetic energy of a monatomic molecule is $0.414 \mathrm{~eV}$ at temperature :

(Use $K_B=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{mol}-\mathrm{K}$)

<p>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:</p> <blockquote> <p>$K_{\text{avg}} = \frac{3}{2} k_B T$</p> </blockquote> <p>Where:</p> <ul> <li>$K_{\text{avg}}$ is the average kinetic energy</li> <li>$k_B$ is the Boltzmann constant, given as $1.38 \times 10^{-23} \, \text{J/K}$</li> <li>$T$ is the temperature in Kelvin</li> </ul> <p>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:</p> <blockquote> <p>$$K_{\text{avg}} = 0.414 \, \text{eV} = 0.414 \times 1.6 \times 10^{-19} \, \text{J}$$</p> </blockquote> <p>Substituting $K_{\text{avg}}$ and $k_B$ into the equation:</p> <blockquote> <p>$\frac{3}{2} k_B T = 0.414 \times 1.6 \times 10^{-19} \, \text{J}$</p> </blockquote> <p>Solving for $T$:</p> <blockquote> <p>$$T = \frac{0.414 \times 1.6 \times 10^{-19} \, \text{J}}{\frac{3}{2} \times 1.38 \times 10^{-23} \, \text{J/K}}$$</p> </blockquote> <p>After performing the calculations:</p> <blockquote> <p>$T \approx 3200 \, \text{K}$</p> </blockquote> <p>Therefore, the temperature at which the average kinetic energy of a monatomic molecule is $0.414 \, \text{eV}$ is approximately 3200 K.</p>