11. -3 points Notes The particle-on-a-ring is a useful model for the motion of e

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11. -3 points Notes The particle-on-a-ring is a useful model for the motion of electrons around a conjugated macrocycle such as octatetrene, for example. Treat the molecule as a circular ring of radius 0.460 nm, with 10 electrons in the conjugated system moving along the perimeter of the ring. Assume based on the Pauli Exclusion Principle that in the ground state of the molecule each state is occupied by two electrons with opposite spins (a) Calculate the energy of an electron in the highest occupied level in joules. (b) Calculate the (absolute magnitude of the) angular momentum of an electron in the highest occuplied level in -s 1-s (c) Calculate the frequency of radiation in Hz that can induce a transition between the highest occupied and lowest unoccupied levels. Hz Submit Answer Save Progress

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Answer:

The motion of electons around a conjugated macrocycle such as octatetrene, ring of radius 0.460nm, with 10 electrons in the conjugated system moving along the perimeter of the ring.

The energy of an electron in the highest occupied level in joules

a)  

The angular momentum states are defined by the quantum number ml = 0, ±1, ±2, · · · , the energy of state ml is given by:

Eml = ml2h2 / 2I

The moment of Innertia (I) is given by:

I = mR2

where m is the mass of an electron (9.11x10-31 kg) so, for a particle of radius of 0.460 nm, Inertia:

I = 9.11x10-31 x 0.46x10-9 = 4.1906x10-40 kg m2

There are 14 electrons with two electrons in each of the lowest 7 energy levels so that the highest occupied states is ml = ±3, assuming h = 1.05457x10-34 Jxs, then

E = 32 (1.05457x10-34)2 / (2x4.1906x10-40)

E3 = 1.1942x10-28 J.

The angular momentum of an electron in the highest occupied level in j-s

b)

The angular momentum is defined by:

lz = mlh

lz = 3x1.05457x10-34 = 3.16371x10-34 Jxs.

The frequency of radiation in Hz that can induce a transition between the highest occupied and lowest unoccupied levels.

c)

The lowest unoccupied energy level is ml = ±4 which has the energy:

E4 = 42 (1.05457x10-34)2 / 2x4.1906x10-40 = 2.1231x10-28 J

DeltaE = E4 - E3 = 2.1231x10-28 - 1.1942x10-28 = 0.9289x10-28 J

And E = h'cf -----> f = E/h'c; where h' = 6.62x10-34 Jxs and c = 3x108 m/s

f = 0.9289x10-28 / 6.63x10-34 x 3x108

f = 0.467x10-3 Hz

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