The energy (in eV) possessed by a neon atom a | vedclass

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(C) Neon is a monoatomic gas,so its degrees of freedom $f = 3$.The average kinetic energy of a monoatomic gas molecule is given by $E = \frac{3}{2} k

Vedclass · Accepted Answer

$3.88 	imes 10^{-2}$

Vedclass · Answer

(C) Neon is a monoatomic gas,so its degrees of freedom $f = 3$.The average kinetic energy of a monoatomic gas molecule is given by $E = \frac{3}{2} k T$.Given temperature $T = 27^{\circ} C = 27 + 273 = 300 \, K$.Boltzmann constant $k = 1.38 	imes 10^{-23} \, J/K$.Energy in Joules $E = \frac{3}{2} 	imes (1.38 	imes 10^{-23}) 	imes 300 = 6.21 	imes 10^{-21} \, J$.To convert energy from Joules to $eV$,divide by $1.6 	imes 10^{-19} \, J/eV$:$E = \frac{6.21 	imes 10^{-21}}{1.6 	imes 10^{-19}} \approx 3.88 	imes 10^{-2} \, eV$.

The energy (in $eV$) possessed by a neon atom at $27^{\circ} C$ is ..............

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