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).

Asked by Pragya Singh | 1 year ago |  105

1 Answer

Solution :-

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

Average thermal energy = (\( \dfrac{3}{2}\)) kT

Where,

k is the Boltzmann constant 

= 1.38 x 10-23 m2 kg s-2 K-1

Hence,

(\( \dfrac{3}{2}\)) kT = (\( \dfrac{3}{2}\)) x 1.38 x 10-23 x 300

On calculation, we get,

= 6.21 x 10-21 J

Therefore, the average thermal energy of a helium atom at room temperature of

 27°C is 6.21 x 10-21 J

 

(ii) On the surface of the sun, T = 6000 K

Average thermal energy = (\( \dfrac{3}{2}\)) kT

= (\( \dfrac{3}{2}\)) x 1.38 x 10-23 x 6000

We get,

= 1.241 x 10-19 J

Therefore, the average thermal energy of a helium atom on the surface of the sun is 1.241 x 10-19 J

(iii) At temperature, T = 107 K

Average thermal energy = (\( \dfrac{3}{2}\)) kT

= (\( \dfrac{3}{2}\)) x 1.38 x 10-23 x 107

We get,

= 2.07 x 10-16 J

Therefore, the average thermal energy of a helium atom at the core of a star is 2.07 x 10-16 J

Answered by Abhisek | 1 year ago

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