Question

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). 11th physics , chapter-13

Answer 1

(1)  here,
 T = 27°C = 27 + 273.15 = 300.15K
we know,
average thermal energy = $\frac{3}{2}k_BT$
$E =\frac{3}{2}*1.38*10^{-38}*300.15\\\\= 6.21*10^{-21}J$

(ii) at the temperature,
       T = 6000K
   average thermal energy = $\frac{3}{2}k_BT$
            = 3/2 *1.38*10^-38*6000
            =1.241*10^-19J

(iii) at the temperature,
         T = 10^7K
average thermal energy = 3/2*1.38*10^-38*10^7
                                         =2.07*10^-16J

 

Answer 2

Hope it helps u.......