The average kinetic energy of a monatomic mol | vedclass

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Question

(B) For a monatomic molecule,the degree of freedom is $f = 3$.The average kinetic energy $K_{avg}$ is given by the formula:$K_{avg} = \frac{3}{2} K_{B

Vedclass · Accepted Answer

$3200$

Vedclass · Answer

(B) For a monatomic molecule,the degree of freedom is $f = 3$.The average kinetic energy $K_{avg}$ is given by the formula:$K_{avg} = \frac{3}{2} K_{B} T$Given:$K_{avg} = 0.414 eV = 0.414 	imes 1.6 	imes 10^{-19} J$$K_{B} = 1.38 	imes 10^{-23} J/K$Substituting the values into the formula:$0.414 	imes 1.6 	imes 10^{-19} = \frac{3}{2} 	imes 1.38 	imes 10^{-23} 	imes T$Solving for $T$:$T = \frac{0.414 	imes 1.6 	imes 10^{-19} 	imes 2}{3 	imes 1.38 	imes 10^{-23}}$$T = \frac{1.3248 	imes 10^{-19}}{4.14 	imes 10^{-23}}$$T = 0.32 	imes 10^4 K = 3200 K$Thus,the temperature is $3200 K$.

The average kinetic energy of a monatomic molecule is $0.414 eV$ at temperature : (Use $K_{B} = 1.38 \times 10^{-23} J/K$) (in $K$)

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