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At what temperature is the rms Speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at `-20^(@)C`? [Atomic mass of Ar = 39.90, He = 4.04]

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Temperature of the helium atom,
`T_(He)=-20^(@)C=273-20=253K`
Atomic mass of argon, `M_(Ar)=39.90`,
Atomic mass of helium, `M_(He)=4.04`
Let, `(V_(rms))_(Ar)` be the rms speed of argon,
Let, `(V_(rms))_(He)` be the rms speed of Helium
The rms speed of argon is given by:
`(V_(rms))_(Ar)=sqrt((3RT_(Ar))/(M_(Ar)))" "...(1)`
`(V_(rms))_(He)=sqrt((3RT_(He))/(M_(He)))" "...(2)`
It is given that: `(V_(rms))_(Ar)=(V_(rms))_(He)`
`sqrt((3RT_(Ar))/(M_(Ar)))=sqrt((3RT_(He))/(M_(He))),(T_(Ar))/(M_(Ar))=(T_(He))/(M_(He))`
`T_(Ar)=(T_(He))/(M_(He))xxM_(Ar)=(253)/(4)xx39.9=2523.675=2.52xx10^(3)K`
Therefore, the temperature of the argon atom is `2.52xx10^(3)K`.
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