How much energy is required to ionize one atom of Al?

[jewilki1]

a. the first ionization energy of aluminum is 578 kJ/mol. How much energy is required to ionize one atom of Al?
I put E= 578 kJ/mol . is this how to start?

b. Write out balanced chemical equation that represents the first ionization of aluminum (you must include the physical states)

c. Calculate the wavelength of a photon that has just enough energy to ionize a single atom of aluminum.

d. how many moles of photons with the wavelength you calculated in part c are required to ionize 1.15 moles of aluminum?

Can you help me set these up. I am studying for a test and we have not been over all of these yet? Thanks.

 

[mrjeffy321]

a. the first ionization energy of aluminum is 578 kJ/mol. How much energy is required to ionize one atom of Al?
I put E= 578 kJ/mol . is this how to start?

the energy given is in terms of moles. You are trying to find the energy for a single atom. So how many atoms in a mole...

b. Write out balanced chemical equation that represents the first ionization of aluminum (you must include the physical states)

Remember, Aluminum like to ionize to +3.

c. Calculate the wavelength of a photon that has just enough energy to ionize a single atom of aluminum.

Using the equation E = H * C / Lambda,
where E is the energy of a wave, H is planks constant, C is the speed of light in a vacuum, and Lamba is the wavelength of the wave, you can solve for the wavelength since you know Energy and H and C are constants.
H = 6.63 E-34 Js/photon
C =3.0 E8 m/s

d. how many moles of photons with the wavelength you calculated in part c are required to ionize 1.15 moles of aluminum?

I think 1 photon will ionize 1 atom assuming the photon is of the correct energy.

 

[Blueshift5]

a. Yes, that is a good start. The first ionization energy of aluminum is 578 kJ/mol, which means that 578 kJ of energy is required to remove one mole (6.022 x 10^23 atoms) of aluminum atoms from their neutral state and form one mole of Al+ ions.

b. The balanced chemical equation for the first ionization of aluminum is: Al(s) --> Al+(g) + e^-. This means that one aluminum atom in its solid state (s) is converted into one aluminum ion in its gaseous state (g) by losing one electron (e^-).

c. To calculate the wavelength of a photon that has just enough energy to ionize a single atom of aluminum, we can use the equation E = hc/λ, where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the photon. Rearranging this equation, we get λ = hc/E. Since we know the energy required to ionize one atom of aluminum (578 kJ/mol), we can convert it to joules (kJ/mol x 1000 J/kJ = 578,000 J/mol) and divide by Avogadro's number (6.022 x 10^23 atoms/mol) to get the energy required to ionize one atom of aluminum (9.598 x 10^-19 J/atom). Plugging this value into the equation, we get λ = (6.626 x 10^-34 J s)(2.998 x 10^8 m/s)/(9.598 x 10^-19 J/atom) = 2.080 x 10^-7 m = 208 nm.

d. To calculate the number of moles of photons required to ionize 1.15 moles of aluminum, we can use the equation n = E/(hc/λ), where n is the number of moles of photons, E is the energy required to ionize 1.15 moles of aluminum (1.15 x 578 kJ/mol = 665.7 kJ), and λ is the wavelength of the photon we calculated in part c (2.080 x 10^-7 m). Plugging in these values, we get n = (