Answer 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.

Answer 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`