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).

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Payal Gupta | Contributor-Level 10

9 months ago

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

$k_{B}$ is Boltzmann constant = 1.38 $* 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

Average thermal energy = $\frac{3}{2} k_{B} T$ = $\frac{3 * 1.38 * 10^{- 23} * 300}{2}$ = 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 = $\frac{3}{2} k_{B} T$ = $\frac{3 * 1.38 * 10^{- 23} * 6000}{2}$ = 1.242 $* 10^{- 19}$ J

Hence, the average

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