1. The lithium dication, Li2+, has an energy level formula analogous to that of

Question

1. The lithium dication, Li2+, has an energy level formula analogous to that of the hydrogen atom, since both species have only one electron. The energy levels of the Li2+ ion are given by the equation E = 11808 kJ/mole n =1, 2, 3, One of the most important transitions for the Li2+ ion involves a jump from the n = 2 to the n = 1 level. E for this transition equals E2 E1, where these two energies are obtained as in Part a. Find the value of E in kJ/mole. Find the wavelength in nm of the line emitted when this transition occurs; use Equation 4 to make the calculation. E = _____________ kJ/mole; l = _____________ nm

Explanation / Answer

Li2+ has the energy level formula analogous to hydrogen atom, since both are 1 electron systems. The energy level formula for hydrogen is

E = -E0/n2 where E = 11808 kJ/mol and n = 1, 2, 3, etc.

We need to evaluate the energy for n = 2 to n = 1 transition. Therefore,

E1 = -11808 kJ/mol/(1)2 and

E2 = -11808 kJ/mol/(2)2

Therefore,

E = E2 – E1 = -11808 kJ/mol/(2)2 – [-11808 kJ/mol/(1)2] = -11808 kJ/mol*[1/(2)2 – 1/(1)2] = -11808 kJ/mol*(1/4 – 1) = -11808 kJ/mol*(-3/4) = 11808 kJ/mol*(3/4) = 8856 kJ/mol (ans).

The above is the energy change for 1 mole of Li2+ cation; we need to calculate the energy change E’ for a single Li2+ cation.

E’ = (8856 kJ/mol)*(1 mole/6.023*1023 molecules) = 1.470*10-20 kJ/molecule = (1.470*10-20 kJ/molecule)*(103 J/1 kJ) = 1.470*10-17 J/molecule.

Again, we know that

E’ = hc/ where h = 6.626*10-34 J.s is the Planck’s constant, c = 3*108 m s-1 is the speed of the wave and = wavelength of the transition.

Plug in values and obtain

1.470*10-17 J/molecule = (6.626*10-34 J.s)*(3*108 m s-1)/

====> 1.470*10-17 J = 1.9878*10-25 J.m/ (we ignore per molecule here).

====> = 1.9878*10-25/(1.470*10-17) m = 1.3522*10-8 m = (1.3522*10-8 m)*(109 nm/1 m) = 13.522 nm (ans).

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