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

For a monatomic gas, the average kinetic energy per molecule is:

\[ KE = \frac{3}{2} k_B T \]

Given \( KE = 0.414 \, \text{eV} \), convert this to joules:

\[ 0.414 \, \text{eV} = 0.414 \times 1.6 \times 10^{-19} \, \text{J} = 6.624 \times 10^{-20} \, \text{J} \]

Now,

\[ 6.624 \times 10^{-20} = \frac{3}{2} \times 1.38 \times 10^{-23} \times T \]

Solving for \( T \):

\[ T = \frac{6.624 \times 10^{-20}}{\left(\frac{3}{2}\right) \times 1.38 \times 10^{-23}} \approx 3200 \, \text{K} \]