At what temperature will the linear kinetic energy of a gas molecule be equal to that of an electron accelerated through a potential difference of 10V?

273 K

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19 \times 10^{3} K

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38.65 \times 10^{3}

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11.3 \times 10^{3} K

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Solution

The correct option is D $11.3 \times 10^{3} K$
$T h e k i n e t i c e n e r g y o f g a s m o l e c u l e = 3 / 2 k T E n e r g y o f t h e a c c e l e r a t e d e l e c t r o n = e V H e n c e e V = 3 / 2 k T F o r t h e g i v e n c a s e k = 1.38 \times 10^{- 23} 1.6 \times 10^{- 19} \times 10 = 1.5 k T T = \frac{1.6 \times 10^{- 19} \times 10 \times 2}{3 k} T = {11.3 \times 10}^{3} K$

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At what temperature will the linear kinetic energy of a gas molecule be equal to that of an electron accelerated through a potential difference of 10V?

The correct option is D 11.3×103KThekineticenergyofgasmolecule=3/2kTEnergyoftheacceleratedelectron=eVHenceeV=3/2kTForthegivencasek=1.38×10−231.6×10−19×10=1.5kTT=1.6×10−19×10×23kT=11.3×103K