Question -
Answer -
(i) At room temperature, T = 270┬аC= 300 K
Average thermal energy = (3 / 2) kT
Where,
k is the Boltzmann constant = 1.38 x 10-23┬аm2┬аkg s-2┬аK-1
Hence,
(3 / 2) kT = (3 / 2) x 1.38 x 10-23┬аx300
On calculation, we get,
= 6.21 x 10-21┬аJ
Therefore, the average thermal energy of a helium atom at roomtemperature of 270┬аC is 6.21 x 10-21┬аJ
(ii) On the surface of the sun, T = 6000 K
Average thermal energy = (3 / 2) kT
= (3 / 2) x 1.38 x 10-23┬аx6000
We get,
= 1.241 x 10-19┬аJ
Therefore, the average thermal energy of a helium atom on thesurface of the sun is 1.241 x 10-19┬аJ
(iii) At temperature, T = 107┬аK
Average thermal energy = (3 / 2) kT
= (3 / 2) x 1.38 x 10-23┬аx107
We get,
= 2.07 x 10-16┬аJ
Therefore, the average thermal energy of a helium atom at thecore of a star is 2.07 x 10-16┬аJ