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Step-by-Step Detailed Solution
Step 1: Identify the Relevant Concepts
To find the temperature at which the average translational kinetic energy of nitrogen ( \text{N}_2 ) gas molecules equals the kinetic energy of an electron accelerated by a 0.1 V potential, we use:
The kinetic energy of a gas molecule (from kinetic theory):
KE_\text{gas} = \frac{3}{2} k_B T
where k_B is the Boltzmann constant and T is the absolute temperature (in kelvin).
The kinetic energy of an electron accelerated through a potential difference V :
KE_\text{electron} = e \, V
where e is the electron charge ( 1.6 \times 10^{-19}\,\text{C} ) and V is the potential difference in volts.
Step 2: Equate the Two Kinetic Energies
Since we want these two energies to be equal,
\frac{3}{2} k_B T = eV.
Step 3: Substitute Known Values
Given:
k_B = 1.38 \times 10^{-23}\,\text{J/K}
e = 1.6 \times 10^{-19}\,\text{C}
V = 0.1\,\text{V}
Substitute these values into the equation:
\frac{3}{2} \times 1.38 \times 10^{-23} \times T = 1.6 \times 10^{-19} \times 0.1.
Step 4: Solve for the Temperature (K)
Rearrange to find T :
T = \frac{1.6 \times 10^{-19} \times 0.1 \times 2}{3 \times 1.38 \times 10^{-23}}.
Upon calculation, this comes out to:
T \approx 773 \,\text{K}.
Step 5: Convert Kelvin to Celsius
To convert from kelvin to degrees Celsius:
T (^\circ \text{C}) = T (K) - 273.
So,
773\,\text{K} - 273 = 500^\circ \text{C}.
Final Answer
The nearest integer Celsius temperature at which the average translational kinetic energy of \text{N}_2 equals the kinetic energy of an electron accelerated through 0.1 V is 500°C.
