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

(i).

Let q be the average thermal energy, so

q= 3 2 k b T…… (1)

Here, k b is the Boltzmann constant which is 1.38× 10 −23   m 2 ⋅kg/ K⋅ s 2 and T is the temperature.

The given temperature is 27 °C.

Substitute the value in the above expression.

q= 3 2 ×1.38× 10 −23 ×( 27+273 ) = 3 2 ×1.38× 10 −23 ×300 =6.2× 10 −21  J

Hence, the average thermal energy of helium atom at room temperature is T=6.2× 10 −21  J.

(ii)

The given temperature of surface of Sun is 6000 K.

Substitute the values in equation (1).

q= 3 2 ×1.38× 10 −23 ×6000 =1.24× 10 −19  J

Hence, 1.24× 10 −19  Jis the average thermal energy of helium atom on the surface of Sun at T=6000 K.

(iii).

The given temperature is T= 10 7  K.

Substitute the values in equation (1).

q= 3 2 ×1.38× 10 −23 × 10 7 =2.1× 10 −16  J

Hence, 2.1× 10 −16  Jis the average thermal energy of helium atom at T= 10 7  K.