Question

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

• A. 3000 K
• B. 3200 K
• C. 1600 K
• D. 1500 K

Answer 1

The Correct Option is B

To determine the temperature at which a monatomic molecule's average kinetic energy equals 0.414 eV, we employ the formula relating average kinetic energy and temperature for a monatomic ideal gas: \(KE_{\text{avg}} = \frac{3}{2} k_B T\). Here, \(KE_{\text{avg}}\) represents the average kinetic energy per molecule, \(k_B\) is the Boltzmann constant (1.38 x 10^-23 J/K), and \(T\) is the absolute temperature in Kelvin.

Initially, the kinetic energy is converted from electronvolts (eV) to joules (J) using the conversion factor 1 eV = 1.602 x 10^-19 J. Thus, 0.414 eV is equivalent to 0.414 x 1.602 x 10^-19 J, which equals 6.633 x 10^-20 J.

Substituting this value into the average kinetic energy equation yields: \(\frac{3}{2} k_B T = 6.633 \times 10^{-20}\).

Solving for \(T\): \(T = \frac{2 \times 6.633 \times 10^{-20}}{3 \times 1.38 \times 10^{-23}} = \frac{1.3266 \times 10^{-19}}{4.14 \times 10^{-23}} = 3204.35\) K.

Rounded to the nearest whole number, the temperature is 3200 K.

Therefore, the answer is 3200 K.