Question

Calculate the energy in eV for the first excited state of an electron in an infinite

potential well of width 2 Å.​

Answer 1

donot know sorry but I think some one else will help u

Answer 2

Answer:

The energy of an electron in the first excited state in an infinite  potential well of width 2 Å is $37.5eV$

Explanation:

• The nth state energy of a particle in an infinite potential well of width L is given by the formula

                                $E_{n} =\frac{n^{2} h^{2} }{8mL^{2} }$

       where,

       $n$-state of the particle

       $m$-the mass of the particle

       $h$-$6.26*10^{-34}Js$Planck's constant

      $L$-width of the potential well

• Joule to electron volt conversion

                            $1J=6.24*10^{18}eV$

From the question, it is given that

the state of the electron is first excited that is $n=2$

the width of the potential well is $L=2*10^{-10}m$

we know that mass of an electron $m=9.1*10^{-31} kg$

substitute these values to get energy as follows

$E_{2} =\frac{4(6.626*10^{-34} )^{2} }{8*9.1*10^{-31} (2*10^{-10}) ^{2} }=0.60*10^{-17}J$

so the energy in eV is $0.60*10^{-17}*6.24*10^{18} eV=37.5eV$