Question

Average K.E. of one single molecule of monoatomic gas is 0.414 eV (of \(\perp\) molecule). What will be is temp = ?

(1) 3000 K

(2) 3200 K

(3) 4292 K

(4) 2500 K

Answer 1

The correct option is (2) 3200 K

\(\frac{f}{2} kT = \frac{3}{2} kT = 0.414 \times 1.6 \times 10^{-19} J\)

T = 3200 K