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).
Solution 1
1) 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
`:. 3/2kT = 3/2 xx 1.38 xx 10^(-38) xx 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.
2) On the surface of the sun, T = 6000 K
Average thermal energy = `3/2 kT`
`= 3/2 xx 1.38 xx 10^(-38) xx 6000`
`= 1.241 xx 10^(-19) J`
Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10–19 J.
3) At temperature, T = 107 K
Average thermal energy `= 3/2 kT`
`= 3/2 xx 1.38 xx 10^(-23) xx 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 × 10–16 J.
Solution 2
1) Here `T = 27 ^@C = 27 + 273 = 300 K`
Average thermal energy = `3/2 kT = 3/2 xx 1.38 xx10^(-23) xx 300 = 6.2 xx 10^(-21) J`
2) At T = 6000 K
Average thermal energy = `3/2 kT = 3/2 xx 1.38 xx 10^(-23) xx 6000 = 1.24 xx 10^(-19) J`
3) At T = 10 million K = `10^7 K`
Average thermal energy = `3/2 kT = 3/2 xx 1.38 xx 10^(-23)xx10^7 = 2.1 xx 10^(-16) J`
