Estimate the average thermal energy of a helium atom at (i) room temperature `(27^(@)C)` (ii) the temperature on the surface of the sun (6000K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star)

Explanation

(i) At room temperature, ` T = 27^(@)C = 300 K`
Average thermal energy `=(3)/(2)kT`
Where k is Boltzmann constant `=1.38 xx 10^(-23) m^2 "kg" s^(-2) K^(-1)`
` therefore(3)/(2)kT=(3/2)xx1.38 xx 10^(-38) xx 300`
`=6.21 xx 10^(-21) J`
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 `=3/2kT`
`3/2xx1.38xx10^(-38)xx6000`
` =1.241 xx 10^(-19)`
Hence, the average thermal energy of a helium atom on the surface of the sun is `1.241 xx 10^(-19) J`
(iii) At temperature, `T = 10^7 K`
Average thermal energy =`3/2kT`
` =3/2xx1.38xx10^(-23)xx10^7`
Hence, the average thermal energy of a helium atom at the core of a star is` 2.07 × 10^(–16) J`.