__Question 13.7__:
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__:**(i)**At room temperature,

*T*= 27°C = 300 K

Average thermal energy = (3/2)kT

Where

*k*is Boltzmann constant = 1.38 × 10

^{–23}m

^{2}kg s

^{–2}K

^{–1}

∴ (3/2)kT = (3/2) X 1.38 X 10

^{-38}X 300

= 6.21 × 10

^{–21}J

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) X 1.38 X 10

^{-38}X 6000

= 1.241 X 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*= 10

^{7}K

Average thermal energy = (3/2)kT

= (3/2) X 1.38 X 10

^{-23}X 10

^{7}

= 2.07 X 10

^{-16}J

Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10

^{–16}J.

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