At what temperature does the average translational KE of a molecule in a gas becomes equal to the K.E. of an electron accelerated through a PD of 3 V?

Question

Prashant Kotian · Accepted Answer

Correct answer: C. average translational k.e. per molecule = 1.5 k_b × t. k.e. of accelerated electron = q × v = 1.6 × 10^-19 c × 3 v = 4.8 × 10^-19 j. equating both: 1.5 × 1.38 × 10^-23 × t = 4.8 × 10^-19 → 2.07 × 10^-23 × t = 4.8 × 10^-19 → t = 23,188 k ≈ 23,200 k.

💡 Solution & Explanation

Practice 22,000+ questions like this