Question

The average translational kinetic energy of a molecule in a gas becomes equal to $0.69\, eV$ at a temperature about
[Boltzmann constant = $1.38 \times 10^{-23} \; J \; K^{-1}$]

• A. $3370 {^{\circ}}C$
• B. $3388 {^{\circ}}C$
• C. $5333 {^{\circ}}C$
• D. $5060 {\circ}C$

Answer 1

Answer: (D)

Given, average translational kinetic energy
$=0.69\, eV =0.69 \times 1.6 \times 10^{-19} \,V$
As we know that,
average translational kinetic energy $=\frac{3}{2} k T$
$0.69 \times 1.6 \times 10^{-19}=\frac{3}{2} \times 1.38 \times 10^{-23} T$
$T=\frac{0.69 \times 1.6 \times 10^{-19} \times 2}{3 \times 1.38 \times 10^{-23}}$
$T=5333 K =5333-273$
$=5060^{\circ} C$