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  • Find the temperature of which the average Kinetic energy of a gas molecule will be equal to the energy of a photon in 6000 A radiation. Given, Boltzmann constant, <img alt="K=1.38 \times 10^{-23} \

Find the temperature of which the average Kinetic energy of a gas molecule will be equal to the energy of a photon in 6000 A radiation. Given, Boltzmann constant, K=1.38 \times 10^{-23} \mathrm{~J}. K^{-1}; Planck's constant, h=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}

Option: 1

1.6


Option: 2

1.6 \times 10


Option: 3

1.6 \times 10^{4} \mathrm{~K}


Option: 4

1.5 \times 10


Answers (1)

best_answer

Let the required temperature T

Average molecular kinetic energy =\frac{3}{2} k T;

Energy of Photon =h v=\frac{h C}{\lambda}

 Here [\left.\lambda=6000 \AA=6000 \times 10^{-10} \mathrm{~m}\right].

According to the question, \frac{3}{2} k T=\frac{h C}{\lambda}

\begin{aligned} & \therefore T=\frac{2}{3} \frac{h c}{1 K}=\frac{2}{3} \times \frac{\left(6.625 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right)}{\left(6000 \times 10^{-10}\right) \times\left(1.38 \times 10^{-23}\right)} \\ & =1.6 \times 10^{4} \mathrm{~K} \end{aligned}


 

Posted by

HARSH KANKARIA

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