Answer:
0.614 nm
Explanation:
Energy of the nth state of one dimensional infinite wall is,
[tex]E=\frac{n^{2} h^{2} }{8mL^{2} }[/tex]
Given the energy to excite an electron from ground state to first excited state is,
[tex]\Delta E=3eV\\\Delta E=3(1.6\times10^{-19})J[/tex]
And the Plank's constant is, [tex]h=6.626\times10^{-34}Js[/tex]
Mass of electron,[tex]m=9.1\times10^{-31}kg[/tex]
Now the energy will of a 1 dimensional infinite wall which excite an electron from ground state to first excited state will be,
[tex]\Delta E=\frac{2^{2} h^{2} }{8mL^{2} } -\frac{1^{2} h^{2} }{8mL^{2} }[/tex]
Put all the variables in above equation and rearrange it for L.
[tex]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[/tex]
Therefore the width of the box is 0.614 nm.
