part A and b calculate the box length, using the particle in th ebox modle for h

Question

part A and b calculate the box length, using the particle in th ebox modle for hexatriene

Calculate the box length, using the particle in the box model for hexatriene, C_6 H_8, assume that the molecule is linear and use the values 135 and 154 pm for C = C and C - C bonds. Express your answer with the appropriate units. Calculate the energy levels of the pi-network in hexatriene, C_6 H_s, using the particle in the box model. The transition is between n = 3 and n = 4. Express your answer with the appropriate units.

Explanation / Answer

Part A

The strucutre of Hexatriene

CH2=CH-CH=CH-CH=CH2

Length of double bond: 135 pm

Length of single bond: 154 pm

Length of Hexatriene = 135 + 154 + 135 + 154 + 135 = 713 pm = 0.713 nm

Length of Hexatriene = 0.713 nm

Part B

E = n2h2/8mL2 (Energy in electron is particle in a box)

E (3) = 9h2/ 8mL2

E(4) = 16h2/8mL2

E = E(3) - E(4) = 9h2/ 8mL2 - 16h2/8mL2 = h2/8mL2 (9-16) = -7h2/8mL2'

E = -7h2/8mL2'

h = 6.625x 10-34Js (planck constant)

m = 9.1 x 10-34 kg (mass of electron)

L = 713x10-12 m (Length of hexatreine)

E = -7(6.625x 10-34Js)2/8(9.1 x 10-34 kg) (713x10-12 m)2

E = 8.3 x 10-16J

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