Question

Find the lowest energy of an electron confined in a Cubical box of each side 2 AU. Also find the temperature at which the average energy of the molecules of a perfect gas would be equal to the lowest energy of the electron (Given-
k = 1.38 x 10^-23 J/°K).​

Answer 1

Answer:

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Answer 2

given side length of a cubical box confining an electron, find its lowest energy

Explanation:

1. Lowest energy of an electron of mass 'm' confined in a cubical box of side length 'a' is given by ,                                                                                               $E=\frac{h}{8ma^2}$                             h->plank's constant=$6.626x10^{-34}\ m^2kg/s$                                                                                    

                                                    m->mass of electron=$9.1x10^{-31}\ kg$                      

    2.given cubical box of side length $a=2\ AU$                                                                          

        AU-Astronomical Unit                  $1 AU=1.5x10^{11}\ m\ (approx.)$                                                  

        hence, we get                               $a= 2AU= 3x10^{11}\ m$                                            

    3.hence, from point 1. we get the lowest energy of electron as,        

         $E=\frac{h^2}{8ma^2} \\E=\frac{(6.626x10^{-34})^2}{8x9.1x10^{-31}x(3x10^{11})^2}\\E=6.7x10^{-61} J$        

    4. average energy of a perfect gas at a temperature 'T'  is given by,

                           $K=\frac{3}{2}{kT}$               k->Boltzmann's constant=$1.38x10^{-23}\ J/K$                                                          

    5. temperature at which average energy of this gas is equal to lowest       energy of electron is given by,

                      $K=\frac{3}{2} kT=E\\T=\frac{2x6.7x10^{-61}}{3x1.38x10^{-23}}\\T=3.23x10^{-38}\ Kelvin$