When a sodium atom absorbs light at 330.2 nm, the 3s electron is promoted to the 4p level. How much energy would be required for ionization once the electron is in the 4p level?
See your other post above.
To calculate the energy required for ionization once the electron is in the 4p level, we need to determine the energy difference between the 4p level and the ionization energy.
First, let's convert the wavelength given (330.2 nm) to energy using the equation:
E = (hc) / λ
where E is the energy, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength in meters.
Converting the wavelength to meters:
330.2 nm = 330.2 x 10^-9 m
Now, let's plug the values into the equation:
E = (6.626 x 10^-34 J s * 3.00 x 10^8 m/s) / (330.2 x 10^-9 m)
Simplifying, we have:
E ≈ 6.011 x 10^-19 J
This energy corresponds to the energy difference between the energy level where the 3s electron was before absorption and the 4p level after absorption.
To calculate the energy required for ionization, we need to subtract the ionization energy of sodium from this value. The ionization energy of sodium is the energy required to remove an electron from a neutral sodium atom.
The ionization energy of sodium is approximately 496 kJ/mol, which can be converted to joules by dividing by Avogadro's number (6.022 x 10^23 mol^-1):
Energy for ionization = (496 x 10^3 J/mol) / (6.022 x 10^23 mol^-1)
Energy for ionization ≈ 8.23 x 10^-19 J
Therefore, the energy required for ionization once the electron is in the 4p level is approximately 8.23 x 10^-19 J.