At what temperature the kinetic energy of a gas molecule is half of the value at 27° C

$\begin{array}{l}\text{ Given }{\mathrm{T}}_1={27}^{\circ }\mathrm{C}=27+273=300\mathrm{K}\\ \mathrm{root} \mathrm{mean} \mathrm{square} \mathrm{velocity} ={\mathrm{V}}_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}\\ Kinetic energy=E=\frac{1}{2}m{{\mathrm{V}}^2}_{\mathrm{rms}}\\ \\ \Rightarrow \mathrm{E}\propto \mathrm{T}\\ \Rightarrow \frac{E_1}{E_2}=\frac{T_1}{T_2}\\ temperatue T_1=27+273=300K;let E_1=E;E_2=\frac{E}{2}; substitute given values \\ \frac{E}{(E/2)}=\frac{300}{T_2}\\ {\mathrm{T}}_2=150\mathrm{K}\\ \text{ Then }{\mathrm{T}}_2=150-273\\ {\mathrm{T}}_2=-{123}^{\circ }\mathrm{C}\end{array}$