Question -
Answer -
(i) At room temperature, T = 270 C= 300 K
Average thermal energy = (3 / 2) kT
Where,
k is the Boltzmann constant = 1.38 x 10-23 m2 kg s-2 K-1
Hence,
(3 / 2) kT = (3 / 2) x 1.38 x 10-23 x300
On calculation, we get,
= 6.21 x 10-21 J
Therefore, the average thermal energy of a helium atom at roomtemperature of 270 C is 6.21 x 10-21 J
(ii) On the surface of the sun, T = 6000 K
Average thermal energy = (3 / 2) kT
= (3 / 2) x 1.38 x 10-23 x6000
We get,
= 1.241 x 10-19 J
Therefore, the average thermal energy of a helium atom on thesurface of the sun is 1.241 x 10-19 J
(iii) At temperature, T = 107 K
Average thermal energy = (3 / 2) kT
= (3 / 2) x 1.38 x 10-23 x107
We get,
= 2.07 x 10-16 J
Therefore, the average thermal energy of a helium atom at thecore of a star is 2.07 x 10-16 J