Question:
Estimate the average thermal energy of a helium atom at (i) room temperature (27 degree 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 × 10–23 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.
Not what you are looking for? Go ahead and submit the question, we will get back to you.
