The ionization energy of a certain element is 472 kJ/mol. However, when the atoms of this element are in the first excited state, the ionization energy is only 166 kJ/mol.

Based on this information, calculate the wavelength of light emitted in a transition from the first excited state to the ground state.

The energy should be 473-166 kJ/mol. I would then convert that to J/atom (use Avogadro's number), then E = hc/wavelength. Solve for wavelength. Check my thinking.

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To calculate the wavelength of light emitted in a transition from the first excited state to the ground state, we can use the equation:

ΔE = hc/λ

Where:
ΔE = Energy difference between the two states
h = Planck's constant (6.626 x 10^-34 J·s or 4.136 x 10^-15 eV·s)
c = Speed of light (2.998 x 10^8 m/s)
λ = Wavelength

First, we need to calculate the energy difference (ΔE) between the first excited state and the ground state. To do that, we subtract the ionization energy of the first excited state from the ionization energy of the ground state:

ΔE = (Ionization energy of the ground state) - (Ionization energy of the first excited state)
= 472 kJ/mol - 166 kJ/mol
= 306 kJ/mol

Now, we convert the energy difference to joules:

ΔE = 306 kJ/mol x (1000 J/1 kJ)
= 306,000 J/mol

Next, we can convert the energy from joules to electron volts (eV):

ΔE = 306,000 J/mol x (1 eV/1.602 x 10^-19 J)
= 1.91 x 10^9 eV/mol

Finally, we can calculate the wavelength using the equation mentioned above. Rearranging the equation, we have:

λ = hc / ΔE

Substituting the values:

λ = (6.626 x 10^-34 J·s) x (2.998 x 10^8 m/s) / (1.91 x 10^9 eV/mol)
≈ 1 x 10^-7 m

Therefore, the wavelength of light emitted in the transition from the first excited state to the ground state is approximately 1 x 10^-7 meters, also known as 100 nanometers.