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Find the temperature at which the averag...

Find the temperature at which the average kinetic energy of a gas molecule will be equal to the energy of a photon to `6000 Angstrom` radiation. Given,Boltzmann constant, `k=1.38 times 10^-23 J.K^-1` Planck's constant, `h=6.625 times 10^-34 J.s`.

Text Solution

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Let the required temperature =T.
Average molecular kinetic energy =`3/2 kT`,
Energy of a photon`=hv=(hc)/lamda`
[Here `lamda=6000lamda=6000times10^-10m`]
According to the question , `3/2 kT=(hc)/lamda`
`thereforeT=2/3(hc)/(lamdak)=2/3times((6.25times10^-34)times(3times10^8))/((6000times10^-10)times(1.38times10^-23))`
`=1.6times10^4K`
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