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gayaneshka [121]
3 years ago
6

A simple cyanine dye molecule has a long central carbon chain, with a total of 8 bonds, between the two nitrogen atoms. One elec

tron per bond is essentially free to move up and down this chain. The potential energy for such an electron can be modeled as a one-dimensional box. The distance between each bond is about 0.14 nm.
What would be the transition corresponding to the lowest energy photon that this molecule could absorb? Would such a photon be in the visible range?
Physics
1 answer:
wariber [46]3 years ago
4 0

Answer:

For one dimensional box

E= n^{2}h^{2}/ 8mL^{2}

For lowest energy transition will be from n =1 to n=2

Hence,

hc/\lambda = 2^{2}h^{2}/ 8mL^{2} - h^{2}/8mL^{2}

h= 6.626 *10-34 J.s

m = 9.31 *10-31 Kg

L= 0.14 * 10-9 m

c= 3 *108 m/s

\lambda = 8*c*m*L^{2} / 3*h

\lambda = 22 nm

Not in visible range

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A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 fro
slavikrds [6]

Answer:

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5 0
3 years ago
At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

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Part b)

KE of SHO at x = A/3

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now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

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now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
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