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

Explanation:

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

Average thermal energy = (3/2)kT

Where k is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–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 (27°C) is 6.21 × 10–21 J.

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

Average thermal energy = (3/2)kT

= (3/2) × 1.38 × 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.

(iii) At temperature, T = 107 K

Average thermal energy = (3/2)kT

= (3/2) × 1.38 × 10-23 × 107

= 2.07 × 10-16 J

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

Answer 2

Answer:

Estimate the average thermal energy of a helium atom at (i) room ... Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10–21 J. Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10–19 J. Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10–16 J.