Answer :

This problem can be solved using the relation between kinetic energy and temperature i.e. average thermal energy = (3/2)kBT per atom.


(i) Average thermal energy of a helium atom at room temperature (27 °C)


At room temperature (T) = 27°C = 300 K


Average thermal energy (ETh) = (3/2)kBT


(Where kB is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1 )


ETh = (3/2)kBT


ETh = (3/2) 1.38 × 10–23 × 300


ETh = 6.21 × 10–21J


The average thermal energy of a helium atom at room temperature is 6.21 × 10–21 J.


(ii) The surface of the sun has T = 6000 K


ETh = (3/2)kBT


ETh = (3/2) × 1.38 × 10-23 × 6000


ETh = 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) The temperature of 10 million


kelvin Average thermal energy = (3/2)kBT


ETh = (3/2) × 1.38 × 10-23 × 107


ETh = 2.07 × 10-16 J


Hence, the average thermal energy of a helium atom (the typical core temperature in the case of a star) is 2.07 × 10–16 J.


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