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 the case of a star).
At room temperature, T=27oC=300K Average thermal energy =(3/2)kT Where, k is Boltzmann constant =1.38×10−23m2kgs−2K−1 ∴(3/2)kT=(3/2)×1.38×10−38×300 =6.21×10−21J Hence, the average thermal energy of a helium atom at room temperature of 27oC is 6.21×10−21J. (ii) On the surface of the sun, T=6000K Average thermal energy =(3/2)kT =(3/2)×1.38×10−38×6000 =1.241×10−19J Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241×10−19J. At temperature, T=107K Average thermal energy =(3/2)kT =(3/2)×1.38×10−23×107 =2.07×10−16J Hence, the average thermal energy of a helium atom at the core of a star is 2.07×10−16J.
