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=3.591 A˚. What is the maximum wavelength (in meters) of radiation capable of ionizing sulfur and producing this effect?
How is this calculated? someone to give the little formula
Look for Lecture S6E4, its the same, the difference is that they are asking you for lamda in the exam instead of the velocity in the lecture. Hope that this helps.
Remember
E(photon) = E(electron)+E(Binding)
Look also for deBroglie wave.
p=h/lambda(meters) ,
E= p^2/2m + 1.659*10^-18,
LambdaMax=hc/E
(what is p?
To find the maximum wavelength of radiation capable of ionizing sulfur and producing the given de Broglie wavelength, we can use the equation:
λ = h / p
where λ is the wavelength, h is the Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the particle.
The momentum of a particle with mass m and velocity v is given by:
p = m * v
For an electron, the mass is approximately 9.109 x 10^-31 kg.
In this case, the de Broglie wavelength (λ) is given as 3.591 Å (Angstroms). To convert it to meters, we use the conversion factor 1 Å = 1 x 10^-10 m.
Therefore, λ = 3.591 x 10^-10 m.
Now, let's calculate the momentum of the photoelectron.
p = h / λ = (6.626 x 10^-34 J·s) / (3.591 x 10^-10 m)
p ≈ 1.843 x 10^-24 kg·m/s
Next, let's calculate the maximum wavelength that can ionize sulfur and produce the given effect.
λ = h / p = (6.626 x 10^-34 J·s) / (1.843 x 10^-24 kg·m/s)
λ ≈ 3.594 x 10^-10 m
Therefore, the maximum wavelength of radiation capable of ionizing sulfur and producing the given effect is approximately 3.594 x 10^-10 meters.