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Asked by Dhanshri Last Modified
Harshit Bhardwaj
(i) At room temperature, T = 27°C = 300 K
Average thermal energy
Where k is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1
∴
= 6.21 × 10–21J
Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10–21 J.
(ii) On the surface of the sun, T = 6000 K
Average thermal energy
= 1.241 × 10–19 J
Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10–19 J.
(iii) At temperature, T = 107 K
Average thermal energy
= 2.07 × 10–16 J
Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10–16 J.
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