QJEEM24Ph2f26: JEE Main 2024 Physics - Class 11 JEEM24Ph2f26 Solution: The average kinetic energy of a monatomic molecule is at…

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.