Question

Electrons-in-a-Box: Consider an electron that can move freely throughout the aromatic orbitals of benzene. Model the electron as a particle in a two-dimensional box 4 Å × 4 Å. a) Compute Δε, the energy change from the ground state to the first excited state, nx = ny = 2. b) Compute the wavelength λ of light that would be absorbed in this transition, if Δε = hc/λ, where h is Planck’s constant and c is the speed of light. c) Will this transition be in the visible part of the electromagnetic spectrum (i.e., is liquid benzene colored or transparent), according to this simple model?

Answer 1

For two dimensional box,

           E = (nx² + ny²) h² / 8ml²

For ground state , nx=1,ny=1

E1= 2h²/8ml²

For first excited state , nx=2,ny=2

E2= 4h²/8ml²

ΔE = E2-E1 = 4h²/8ml² - 2h²/8ml²

ΔE = 2h²/8ml² = h²/4ml²

m= mass of electron = 9.1 x 10^-31 kg

l = length of box = 4 Ao = 4 x 10^-10 m

a) ΔE = h²/4ml²

b) ΔE = h²/4ml²

hc/λ =  h²/4ml²

λ = c x 4ml² / h

   = 3 x10⁸ m/s x 4 x 9.1 x 10^-31 kg x (4 x 10^-10 m)² / 6.626 x 10^-34 J.s

= 263.6 x 10^-9 m

= 263.6 nm

λ = 263.6 nm

c) This transition not in the visible part of the electromagnetic spectrum.

       visible region = 380 nm - 700 nm