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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).

0 4 Views | Posted 5 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    5 months ago

    13.7 (i) At room temperature, T = 27 ? = 300 K

    kB is Boltzmann constant = 1.38 *10-23 m2kgs-2K-1

    Average thermal energy = 32kBT = 3*1.38*10-23*3002 = 6.21 *10-21 J

    Hence, the average thermal energy of a helium atom at room temperature is 6.21 *10-21 J

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

    Hence average thermal energy = 32kBT = 3*1.38*10-23*60002 = 1.242 *10-19 J

    Hence, the average thermal energy of a helium atom on the surface of the Sun is 1.242 *10-19 J

    (iii) Inside the core of a star, T = 107 K

    Hence average thermal energy = 32kBT = 3*1.38*10-23*1072 = 2.07 *10-16 J

    Hence, the average thermal energy of a helium atom

    ...more

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