The average translational kinetic energy of a molecule in a gas becomes equal to \(0.69 \mathrm{eV}\) at…
The average translational kinetic energy of a molecule in a gas becomes equal to \(0.69 \mathrm{eV}\) at temperature about, [Boltzmann's constant \(=138 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\) ]
\(3370^{\circ} \mathrm{C}\)
\(3388^{\circ} \mathrm{C}\)
\(5333^{\circ} \mathrm{C}\)
\(5060^{\circ} \mathrm{C}\)
Solution
Given, average translational kinetic energy
\(=0.69 \mathrm{eV}=0.69 \times 1.6 \times 10^{-19} \mathrm{~V}\)
As we know that, average translational kinetic energy \(=\frac{3}{2} k T\)
\(\begin{aligned}
& 0.69 \times 1.6 \times 10^{-19}=\frac{3}{2} \times 1.38 \times 10^{-23} T \\
& T=\frac{0.69 \times 1.6 \times 10^{-19} \times 2}{3 \times 1.38 \times 10^{-23}} \\
& T=5333 \mathrm{~K}=5333-273=5060^{\circ} \mathrm{C}
\end{aligned}\)