CH_2 CH-CH-CH-CH-CH_2 molecule contains 6 carbon atoms with conjugated pi-electr

Question

CH_2 CH-CH-CH-CH-CH_2 molecule contains 6 carbon atoms with conjugated pi-electrons. These pi-electrons can be approximately treated as free particles in a 1D-box with total length L = 0.84 nm. (a) Calculate the energy E_n at n = 4 and n = 5, respectively. (b) When one electron has a transition from n = 4 to n = 5 state by absorption of radiation, what is the energy required? (c) What is the radiation frequency corresponding to this energy absorption? (d) What is the wave-length of this radiation? (Plank's constant h = 6.626 times 10^-34 J s; electron mass m = 9.1 times 10^-31 kg; radiation speed C = 3 times 10^8 m/s; 1 m = 10^9 nm)

Explanation / Answer

a)

Apply Rydberg Formula

E = R*(1/nf^2)

R = -2.178*10^-18 J

Ni = initial stage/level

E = Energy per unit (i.e. J/photon)

E = (-2.178*10^-18)*(1/4^2)

E = 1.3612*10^-19 J/electron

b)

E = ( 2.178*10^-18)*(1/5^2)

E = 8.71*10^-20 J/electron

B)

1 electron goes for 4 to 5... apply energ:

dE = E2-E1 =  1.3612*10^-19 - 8.71*10^-20= 4.902*10^-20 J required per electron

c)

find frequency

h = Planck Constant = 6.626*10^-34 J s

E = energy per particle J/photon

E = hv

v = E/h = (4.902*10^-20 ) /(6.626*10^-34) = 7.3981*10^13 Hz

D)

find WL:

WL = c/v

WL = (3*10^8)/(7.3981*10^13) = 0.0000040550

WL = 4*10^-6 m

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