Answer:
Step-by-step explanation:
To calculate the temperature at which the average kinetic energy (KE) of a gas molecule is equal to the KE of an electron accelerated through 1 volt, we can use the equation for the average kinetic energy of a gas molecule:
KE_gas = (3/2) * k * T_gas
where KE_gas is the average kinetic energy of a gas molecule, k is the Boltzmann constant (approximately 1.38 × 10^-23 J/K), and T_gas is the temperature of the gas in Kelvin.
The kinetic energy of an electron accelerated through a potential difference of 1 volt is given by:
KE_electron = q * V
where KE_electron is the kinetic energy of the electron, q is the charge of the electron (approximately 1.6 × 10^-19 C), and V is the voltage (1 volt in this case).
Setting the two equations equal to each other, we have:
(3/2) * k * T_gas = q * V
Rearranging the equation to solve for T_gas, we get:
T_gas = (2 * q * V) / (3 * k)
Substituting the values for q, V, and k into the equation, we can calculate the temperature:
T_gas = (2 * 1.6 × 10^-19 C * 1 V) / (3 * 1.38 × 10^-23 J/K)
T_gas ≈ 2.32 × 10^4 K
Therefore, the temperature at which the average kinetic energy of a gas molecule is equal to the kinetic energy of an electron accelerated through 1 volt is approximately 23,200 Kelvin.