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Home » Estimate the average thermal energy of a helium atom at(i) room temperature ,(ii) the temperature on the surface of the Sun (6000 K)
Estimate the average thermal energy of a helium atom at
(i) room temperature \left(27^{\circ} \mathrm{C}\right),
(ii) the temperature on the surface of the Sun (6000 K)
CBSE | Class 11 | Kinetic Theory | NCERT | Physics
Estimate the average thermal energy of a helium atom at
(i) room temperature \left(27^{\circ} \mathrm{C}\right),
(ii) the temperature on the surface of the Sun (6000 K)

(i) At room temperature the temperature is T=27^{\circ} \mathrm{C}=300 \mathrm{~K}

Average thermal energy will be =(3 / 2) \mathrm{kT}

Where,

k is the Boltzmann constant having value 1.38 \times 10^{-23} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-2} \mathrm{~K}^{-1}

So,

(3 / 2) \mathrm{kT}=(3 / 2) \times 1.38 \times 10^{-23} \times 300

On calculating, we get,

As a result, 27^{\circ} \mathrm{C} is 6.21 \times 10^{-21} \mathrm{~J} the average thermal energy of a helium atom at room temperature.

(ii) On the surface of the sun temperature is T=6000 \mathrm{~K}

=1.241\times10^{-19}J

As a result, the average thermal energy of a helium atom on the surface of the sun is 1.241 \times 10^{-19} \mathrm{~J}.

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