Question

It takes 3.0 eV of energy to excite an electron in a 1-dimensional infinite well from the ground state to the first excited state. What is the width L of the box?

Answer 1

Answer:

0.614 nm

Explanation:

Energy of the nth state of one dimensional infinite wall is,

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

Given the energy to excite an electron from ground state to first excited state is,

$\Delta E=3eV\\\Delta E=3(1.6\times10^{-19})J$

And the Plank's constant is, $h=6.626\times10^{-34}Js$

Mass of electron,$m=9.1\times10^{-31}kg$

Now the energy will of a 1 dimensional infinite wall which excite an electron from ground state to first excited state will be,

$\Delta E=\frac{2^{2} h^{2} }{8mL^{2} } -\frac{1^{2} h^{2} }{8mL^{2} }$

Put all the variables in above equation and rearrange it for L.

$L^{2} =\frac{3((6.626\times10^{-34}Js)^{2} )}{8\times9.1\times10^{-31}kg\times3(1.6\times10^{-19})J} \\L^{2} =0.376922012\times10^{-18} m^{2}\\ L=0.6139\times10^{-9}m\\ L=0.614nm$

Therefore the width of the box is 0.614 nm.