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 = [tex]\frac{h^2}{8me^2}(n_4^2-n_3^2)[/tex]
ΔE = hv = [tex]\frac{h^2}{8me^2}(4^2-3^2)[/tex]
ΔE = hv = [tex]\frac{h^2}{8me^2}(16-9)[/tex]
ΔE = hv = [tex]\frac{7*h^2}{8me^2}[/tex]
where h = planck constant = [tex]6.626*10^{-34}[/tex] J.s
mass (m) = [tex]9.11*10^{-11}[/tex]
e = 0.904 nm = [tex]0.9*10^{-9} m[/tex]
hv = [tex]\frac{7*h^2}{8me^2}[/tex]
hv = [tex]\frac{7*(6.6246*10^{-34})^2}{8*9.11*10^{-31}*(0.9*10^{-9})^2}[/tex]
hv = [tex]\frac{3.07197276*10^{-66}}{5.90328*10^{-48}}[/tex]
hv = [tex]5.20384051*10^{-19}[/tex] J
v = [tex]\frac{5.20384051*10^{-19}}{h}[/tex]
v = [tex]\frac{5.20384051*10^{-19}J}{6.626*10^{-34}J.s}[/tex]
v = [tex]7.85366814 *10^{14}s^{-1[/tex]
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
