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

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(C) Temperature,$T = 77^{\circ} C = 273 + 77 = 350 \,K$. Boltzmann constant,$K_{B} = 1.38 	imes 10^{-23} \,J/K$. The average kinetic energy of a mono

Vedclass · Accepted Answer

$4.52 	imes 10^{-2}$

Vedclass · Answer

(C) Temperature,$T = 77^{\circ} C = 273 + 77 = 350 \,K$. Boltzmann constant,$K_{B} = 1.38 	imes 10^{-23} \,J/K$. The average kinetic energy of a monoatomic gas atom is given by $E = \frac{3}{2} K_{B} T$. Substituting the values: $E = \frac{3}{2} 	imes 1.38 	imes 10^{-23} 	imes 350 \,J$. To convert the energy from Joules to electron-volts $(eV)$,divide by the charge of an electron $(1.6 	imes 10^{-19} \,C)$: $E(eV) = \frac{1.5 	imes 1.38 	imes 350 	imes 10^{-23}}{1.6 	imes 10^{-19}} \,eV$. $E(eV) = \frac{724.5 	imes 10^{-23}}{1.6 	imes 10^{-19}} \,eV$. $E(eV) \approx 4.52 	imes 10^{-2} \,eV$.

The energy (in $eV$) possessed by a neon atom at $77^{\circ} C$ is (Boltzmann constant,$K_{B} = 1.38 \times 10^{-23} \,J \,K^{-1}$)

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