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Photons and Ionization Energy for calcium

Submitted by biogirl on Thu, 10/02/2008 - 19:47

Hi everyone.  I am stuck on a couple of questions from my lab class and I'd really appreciate it if someone could walk me through them.  I have 2 questions, but I figured I'd save space on the front page and post them both here.  Help on either one or both would be great!  I really need to understand this stuff.  (1b and 2 are where I have diffculty)

1) One of the wavelengths of light emitted by excited calcium atoms is 393 nm.

a) Calculate the energy of the photons emitted by one emitted photon, in joules.

I used E=(hc)/wavelength to come up with 5.06 x 10^-19 J.

b)Calculate the energy of the photons emitted by one mole of atoms, in kJ and in    kcal.  (Remember that 1 cal=4.184 J and 6.022 x 10^23 atoms = 1 mol)

2) The energy needed to remove the outermost electron completely from an atom is called its first ionization energy.  In terms of the Bohr model, ionization can be considered as a process in which the election moves from the lowest energy level to an "orbit" of infinite radius.  Calculate the ionization energy (in kJ/mol) for hydrogen atoms.

If one photon produces 5.06 x 10^-19 J. then a mole of photons would produce

    5.06 x 10-19 J  x  6.022 x 1023 photons  =  ?
        1 photon                  1 mole

For number 2 calculate the energy of an electron at the 1st energy level the the subtract the energy of the electron at the infinite energy level and find the difference. The equation often used for electron energy is

    E  =  2.178 x 10-18 Joules x (Z2 / n2)
Where Z is the atomic number and n is the energy level. When the electron is at an infinite distance from the nucleus n become infinity so E = 0.

Hope this helps,


Yes, it helps a lot.  Thanks so much, Otis.