JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 24)
The average translational kinetic energy of N2 gas molecules at .............$$^\circ$$C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given kB = 1.38 $$\times$$ 10$$-$$23 J/K) (Fill the nearest integer).
Answer
500
Explanation
Given, the average translational kinetic energy of dinitrogen (N2) = Kinetic energy of an electron .... (i)
Translational kinetic energy of dinitrogen (N2)
$$KE = {3 \over 2}{K_B}T$$
Here, T = temperature of the gas,
and KB = Boltzmann constant.
Kinetic energy of an electron = eV
Given, the potential differential of an electron, V = 0.1 V
Substituting the values in the Eq. (i), we get
$${3 \over 2}{K_B}T = eV$$
$$ \Rightarrow {3 \over 2} \times 1.38 \times {10^{ - 23}} \times T = 1.6 \times {10^{ - 19}} \times (0.1)$$
$$T = 773K = 773 - 273^\circ C = 500^\circ C$$
Translational kinetic energy of dinitrogen (N2)
$$KE = {3 \over 2}{K_B}T$$
Here, T = temperature of the gas,
and KB = Boltzmann constant.
Kinetic energy of an electron = eV
Given, the potential differential of an electron, V = 0.1 V
Substituting the values in the Eq. (i), we get
$${3 \over 2}{K_B}T = eV$$
$$ \Rightarrow {3 \over 2} \times 1.38 \times {10^{ - 23}} \times T = 1.6 \times {10^{ - 19}} \times (0.1)$$
$$T = 773K = 773 - 273^\circ C = 500^\circ C$$
Comments (0)
