HW Key _4 S problems part c

HW Key _4 S problems part c - S15 When metallic sodium is...

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S15. When metallic sodium is dissolved in liquid sodium chloride, electrons are released into the liquid. These dissolved electrons absorb light with a wavelength near 800 nm. Suppose we treat the positive ions surrounding an electron very crudely as defining a three-dimensional cubic box of edge L , and we assume that the absorbed light excites the electron from its ground state to the first excited state. Calculate the edge length L within this simple model. For this problem we are treating the solvated electron as a particle in a box. Note there is only one electron in the box, which has sides of length L, which is what we are trying to find. What we want to find is the size of the box, such that a photon with a wavelength of 800 nm has enough energy to excite the electron from the ground state (its lowest energy state) to the first excited state, the state next higher in energy. As we have learned, the energy of the photon must be the same as the energy difference between the two states when absorbed or emitted. The energy of light is given by E = h ν = hc λ To find the energy difference between the two states, we need to consider the particle in a box model. For a 3 dimensional particle in a box model, the energy of a given state can be written as E = h 2 8 m n x 2 L x 2 + n y 2 L y 2 + n z 2 L z 2 Because the length of each side is equal, L x = L y = L z = L , we can replace them with L and factor it out from our energy expression to give us E = h 2 8 mL 2 n x 2 + n y 2 + n z 2 ( ) The smallest value this energy can have is when the quantum numbers, n, have their lowest possible values. Since the smallest value the particle-in-a-box quantum numbers can have is n =1, the ground state must be when n x =n y =n z =1. The ground state then has an energy of E 111 = h 2 8 mL 2 1 2 + 1 2 + 1 2 ( ) = 3 h 2 8 mL 2 The first excited state is the next highest energy state. This state corresponds to one of the quantum numbers changing to n= 2, while the other two remain at a value of 1, but which one changes? In the problem above, the order of the energy states was dependant on which of the quantum numbers changed. That was because the box had different lengths of side; however, in this case where all the sides are the same length, it doesn't matter which of the three quantum numbers change, in fact there are three different states that all have the same energy. These states are designated by n x = 2, n y = 1, n z = 1; n x = 1, n y = 2, n z = 1; n x = 1, n y = 1, n z = 2 or in a shorter notation (211), (121), and (112). Because we have three states that all
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have the same energy, we say these states are degenerate. The energy difference between all of these states and the ground state is the same. We can find this difference according to Δ E = E 211 E
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HW Key _4 S problems part c - S15 When metallic sodium is...

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