Question

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).

Answer 1

At room temperature, T = 27°C = 300 K 

Average thermal energy  \(=\frac{3}{2}\,KT\)

Where k is Boltzmann constant = 1.38 × 10 ^{-23 m}2 kg s^-2 K^-1

∴ \(\frac{3}{2}\,KT=\frac{3}{2}×1.30×10^{-38}×300\)

= 6.21 × 10^–21J 

Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21×10^–21 J. 

On the surface of the sun, T = 6000 K 

Average thermal energy  \(=\frac{3}{2}\,KT\)

\(\frac{3}{2}×1.30×10^{-38}×6000\)

= 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. 

At temperature, T = 10⁷ K  

Average thermal energy  \(=\frac{3}{2}\,KT\)

\(\frac{3}{2}×1.38×10^{-23}×10^7\)

= 2.07 × 10^–16 J 

Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10^-16J.