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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). - Physics

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).

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Solution 1

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

Average thermal energy = `3/2 kT`

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

`:. 3/2kT = 3/2 xx 1.38 xx 10^(-38) xx 300`

= 6.21 × 10–21J

Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10–21 J.

2) On the surface of the sun, T = 6000 K

Average thermal energy = `3/2 kT`

`= 3/2 xx 1.38 xx 10^(-38) xx 6000`

`= 1.241 xx 10^(-19) J`

Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10–19 J.

3) At temperature, T = 107 K

Average thermal energy  `= 3/2 kT`

`= 3/2 xx 1.38 xx 10^(-23) xx 10^7`

= 2.07 × 10–16 J

Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10–16 J.

Solution 2

1) Here `T = 27 ^@C = 27 + 273 = 300 K`

Average thermal energy = `3/2 kT = 3/2 xx 1.38 xx10^(-23) xx 300 = 6.2 xx 10^(-21) J`

2) At T = 6000 K

Average thermal energy = `3/2 kT = 3/2 xx 1.38 xx 10^(-23) xx 6000 = 1.24 xx 10^(-19) J`

3) At T = 10 million K  = `10^7 K`

Average thermal energy = `3/2 kT = 3/2 xx 1.38 xx 10^(-23)xx10^7 = 2.1 xx 10^(-16) J`

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 13 Kinetic Theory
Q 7 | Page 334
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