The average kinetic energy of a monatomic molecule is 0.414 eV at temperature: (Use \( k_B = 1.38 \times 10^{-23} \, \text{J/mol-K} \))
- 3000 K
- 3200 K
- 1600 K
- 1500 K
The Correct Option is B
Approach Solution - 1
To find the temperature at which the average kinetic energy of a monatomic molecule is 0.414 eV, we can use the formula that relates the average kinetic energy of a gas molecule to the temperature. The average kinetic energy per molecule for a monatomic ideal gas is given by the equation:
\(KE_{\text{avg}} = \frac{3}{2} k_B T\)
Where:
- \(KE_{\text{avg}}\) is the average kinetic energy per molecule.
- \(k_B\) is the Boltzmann constant, given as \(1.38 \times 10^{-23} \, \text{J/K}\).
- \(T\) is the absolute temperature in Kelvin.
First, convert the given kinetic energy from electronvolts (eV) to joules (J):
\(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\)
Therefore,
\(0.414 \, \text{eV} = 0.414 \times 1.602 \times 10^{-19} \, \text{J} = 6.633 \times 10^{-20} \, \text{J}\)
Now, substitute this value into the equation for average kinetic energy:
\(\frac{3}{2} k_B T = 6.633 \times 10^{-20}\)
Solving for \(T\):
\(T = \frac{2 \times 6.633 \times 10^{-20}}{3 \times 1.38 \times 10^{-23}}\)
\(T = \frac{1.3266 \times 10^{-19}}{4.14 \times 10^{-23}}\)
\(T = 3204.35 \, \text{K}\)
Rounded to the nearest whole number, this gives \(T = 3200 \, \text{K}\).
Thus, the correct answer is 3200 K.
Approach Solution -2
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 \( T \):
\[ T = \frac{6.624 \times 10^{-20}}{\left(\frac{3}{2}\right) \times 1.38 \times 10^{-23}} \approx 3200 \, \text{K} \]
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