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).
Answers
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.
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.