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 m2 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.