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Kinetic Theory

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Question : 7 of 14
Marks: +1, -0
Estimate the average thermal energy of a helium atom at (i) room temperature (27°C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in case of a star).
Solution:  
The average kinetic energy of the gas at a temperature T
E=32kTE = \frac{3}{2} k T
where k=1.38×10−23 JK−1k = 1.38 \times 10^{-23} \ \mathrm{JK}^{-1}
T = Temperature in kelvin
(i) T=27∘C=27+273T = 27^{\circ}\mathrm{C} = 27 + 273 =300 K= 300 \ \mathrm{K}
E=32×1.38×10−23×300E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 =6.21×10−21 J= 6.21 \times 10^{-21} \ \mathrm{J}
(ii) T=6000 KT = 6000 \ \mathrm{K}
E=32×1.38×10−23×6000E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 6000 =1.242×10−19 J= 1.242 \times 10^{-19} \ \mathrm{J}
(iii) T=10×106 K=107 KT = 10 \times 10^{6} \ \mathrm{K} = 10^{7} \ \mathrm{K}
E=32×1.38×10−23×107E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 10^{7} =2.07×10−16 J= 2.07 \times 10^{-16} \ \mathrm{J}
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