Photons, wavelengths, and ionization energy

[confusedbyphysics]

This is a post-lab question for my chemistry lab but it is kind of physics too so I thought I'd post it here (the other sciences sections don't much response).. It's confusing because the questions are supposed to go along with what we are learning in the chemistry lecture, but we're behind in the lecture, so I'm trying to teach myself how to do this since we haven't learned it yet!

Here's the problem:

"The ionization energy of Li is 520 kJ/electron mole. What is the wavelength of nm of a photon having sufficient energy to remove an electron from Li? In what portion of the electromagnetic spectrum would you find such a photon?"

My work:

520 kJ = 520000 J

520000 J/mol X 1 mol / 6.02 X 10^23 photons = 8.6379 X 10^-19 J / photons

Energy = h (constant) X v (frequency)

v = E/h

v = (8.6379 X 10^-19 J/photons) / (6.626 X 10 ^-34 J/s)

v = 1.3036 X 10 15 /s

wavelength (w) = speed of light (c)/ v

w = (3 X 10^8 m/s) / (1.3036 X 10^15)

w = 2.301 m /photon X 10^9 nm/m = 230 nm

230 nm = ultraviolent rays

Does this look right? If someone could help me Id soooo appreciate it! (I need to make sure I did this right or I'll miss the next problem too because they're related)

 

[lightgrav]

Looks good ... equations used correctly.
I didn't check your calculator punching,
but your final answer is about right.

 

[quicknote]

I would say that your calculations and reasoning seem to be correct. The ionization energy of an element is the minimum amount of energy required to remove an electron from an atom in its ground state. In this case, the ionization energy of Li is 520 kJ/electron mole, meaning that 520 kJ of energy is needed to remove one mole of electrons from Li atoms.

To find the wavelength of a photon with enough energy to remove an electron from Li, we can use the equation E = hv, where E is the energy of the photon, h is Planck's constant, and v is the frequency of the photon. We can then rearrange this equation to solve for v, which gives us the frequency of the photon.

Next, we can use the equation c = fλ, where c is the speed of light, f is the frequency, and λ is the wavelength, to find the wavelength of the photon.

Based on your calculations, the wavelength of the photon is 230 nm, which falls in the ultraviolet region of the electromagnetic spectrum. This makes sense, as higher energy photons (such as those in the ultraviolet region) are needed to remove an electron from an atom.

Overall, your work and reasoning seem to be correct. Just make sure to double check your calculations and units to ensure accuracy. Keep up the good work!