Question

We can model the 1,3,5-hexatriene as a box of length 0.904 nm, with 6 electrons in the first three energy levels. What is the frequency of an electronic transition from the highest-occupied orbital to the lowest-unoccupied orbital?

Answer 1

Answer:

v = 7.85 × 10¹⁴ Hz

Explanation:

The first three energy level can be represented as follows:

⇅ ------> n₄ = 4

⇅ ------> n₃ = 3

⇅ ------> n₂  = 0

ΔE = hv = $\frac{h^2}{8me^2}(n_4^2-n_3^2)$

ΔE = hv = $\frac{h^2}{8me^2}(4^2-3^2)$

ΔE = hv = $\frac{h^2}{8me^2}(16-9)$

ΔE = hv = $\frac{7*h^2}{8me^2}$

where h = planck constant = $6.626*10^{-34}$ J.s

mass (m) = $9.11*10^{-11}$

e = 0.904 nm = $0.9*10^{-9} m$

hv = $\frac{7*h^2}{8me^2}$

hv = $\frac{7*(6.6246*10^{-34})^2}{8*9.11*10^{-31}*(0.9*10^{-9})^2}$

hv =  $\frac{3.07197276*10^{-66}}{5.90328*10^{-48}}$

hv = $5.20384051*10^{-19}$ J

v = $\frac{5.20384051*10^{-19}}{h}$

v = $\frac{5.20384051*10^{-19}J}{6.626*10^{-34}J.s}$

v =  $7.85366814 *10^{14}s^{-1$

v = 7.85 × 10¹⁴ Hz     (since s⁻¹ is equivalent to 1 Hz)

Thus, the frequency of an electronic transition from the highest-occupied orbital to the lowest-unoccupied orbital = 7.85 × 10¹⁴ Hz