Sulfur vapor is analyzed by photoelectron spectroscopy (PES). Measurements determine that photoelectrons associated with the 1st ionization energy of sulfur move with de Broglie wavelength λ=5.591 A˚. What is the maximum wavelength (in meters) of radiation capable of ionizing sulfur and producing this effect?
What is the equation
Don't answer this question until monday. It is for Edx Mit midterm exam class.
equation please
did you get an answer?
How is this calculated? someone to give the little formula
p=h/lambda(meters) ,
E= p^2/2m + 1.659*10^-18,
LambdaMax=hc/E
To determine the maximum wavelength of radiation capable of ionizing sulfur and producing the given de Broglie wavelength, we need to use the relationship between the wavelength (λ) and the energy (E) of a photon.
The energy of a photon can be calculated using the equation:
E = hc / λ
where h is Planck's constant (6.626 x 10^-34 J·s) and c is the speed of light (3.00 x 10^8 m/s).
Now, since we are interested in finding the maximum wavelength (λ_max), we can rearrange the equation to solve for λ_max:
λ_max = hc / E
The energy E can be calculated using the equation:
E = -R∞ / n^2
where R∞ is the Rydberg constant (2.18 x 10^-18 J) and n is the principal quantum number for the energy level of the electron.
In this case, we are given that the de Broglie wavelength (λ) is equal to 5.591 Å (angstroms). To convert angstroms to meters, we use the conversion factor 1 Å = 1 x 10^-10 m.
So, the de Broglie wavelength in meters (λ) is:
λ = 5.591 x 10^-10 m
Now, we can substitute this value into the equation for the energy (E) and calculate it:
E = -R∞ / n^2
To find the maximum wavelength, we need to consider ionization, which occurs when n = ∞. Substituting n = ∞ into the energy equation:
E = -R∞ / ∞^2
E = 0 J
Since the energy is zero, we can substitute this value into the equation for λ_max:
λ_max = hc / E
λ_max = hc / 0
λ_max = undefined
Therefore, the maximum wavelength capable of ionizing sulfur and producing the given de Broglie wavelength is undefined.