Question

An electron is confined in a 3-dimensional rectangular box (V = 0 inside and V = [infinity] outside) with sides L = 4 nm, 4 nm and 5 nm. What minimum energy E must a photon have in order to excite the electron out of its ground state. (The electron absorbs the photon.)

Answer 1

0.045 eV

Step-by-step explanation:

The energy of the electron which is confined in the three dimensional rectangular box is,

`E=(h^(2) )/(8m)( ((n_(x) )/(L_(x) )) ^(2)-((n_(y) )/(L_(y) )) ^(2)-((n_(z) )/(L_(z) )) ^(2))`

Here for the ground state all the values of n will be,

`n_(x)= n_(y)=n_(z)=1`

And according to question,

`L_(x) =4nm=4* 10^(-9)m`

`L_(y) =4nm=4* 10^(-9)m`

`L_(z) =5nm=5* 10^(-9)m`

Therefore, ground state energy will be,

`E_(1) =((6.626* 10^(-34)Js) )/(8(9.1* 10^(-31)kg) ) ((1)/((4* 10^(-9)m)^(2) ) -(1)/((4* 10^(-9)m)^(2) ) }-(1)/((5* 10^(-9)m)^(2) ))\\ E_(1) =9.94* 10^(-21) J((1eV)/(1.6* 10^(-19) ))\\ E_(1) =0.062eV`

Now the minimum energy required to excite the electron from its ground state is,

`E=E_(2)-E_(1)`

So for first excited state there will be 3 possibilities are (1,1,2), (2,1,1), (1,2,1). So in these three cases minimum energy case is (1,1,2)

So minimum energy for first excited state is,

`E_(1) =((6.626* 10^(-34)Js) )/(8(9.1* 10^(-31)kg) ) ((1)/((4* 10^(-9)m)^(2) ) -(1)/((4* 10^(-9)m)^(2) ) }-(2)/((5* 10^(-9)m)^(2) ))\\ E_(1) =0.1073eV`

Sop the minimum energy will be,

`E=0.1073eV-0.062eV\\E=0.045eV`

Therefore minimum energy required to excite electron from its ground state is 0.045 eV