Question 7

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

(i) At room temperature, T= 27°C = 300 K

Average thermal energy = 3/2 kT

Where k is Boltzmann constant = 1.38 × 10-23m2 kg s-2K-1

∴ 3/2 kT = 3/2 x 1.38 x 10-38 x 300

= 6.21 × 10-21J

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

 

(ii) On the surface of the sun, T= 6000 K

Average thermal energy  =  3/2 kT

= 3/2 x 1.38 x 10-38 x 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 .

 

(iii) At temperature, T= 107K

Average thermal energy = 3/2 kT

= 3/2 x 1.38 x 10-38 x 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.

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