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

Answer:

0.045 eV

Explanation:

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

$E=\frac{h^{2} }{8m}( (\frac{n_{x} }{L_{x} }) ^{2}-(\frac{n_{y} }{L_{y} }) ^{2}-(\frac{n_{z} }{L_{z} }) ^{2})$

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

$n_{x}= n_{y}=n_{z}=1$

And according to question,

$L_{x} =4nm=4\times 10^{-9}m$

$L_{y} =4nm=4\times 10^{-9}m$

$L_{z} =5nm=5\times 10^{-9}m$

Therefore, ground state energy will be,

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

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,

$E_{1} =\frac{(6.626\times 10^{-34}Js) }{8(9.1\times 10^{-31}kg) } (\frac{1}{(4\times 10^{-9}m)^{2} } -\frac{1}{(4\times 10^{-9}m)^{2} } }-\frac{2}{(5\times 10^{-9}m)^{2} })\\ E_{1} =0.1073eV$

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