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
p=h/lambda(meters) ,
E= p^2/2m + 1.659*10^-18,
LambdaMax=hc/E
To find the maximum wavelength of radiation capable of ionizing sulfur and producing the given effect, we need to use the equation:
λ = h / p
Where:
λ is the wavelength of the photon
h is the Planck's constant (6.626 x 10^-34 J·s)
p is the momentum of the photon
The momentum of a photon is given by:
p = m * v
Where:
m is the mass of the particle
v is the velocity of the particle
The velocity of the particle can be calculated using the de Broglie wavelength:
v = λ * f
Where:
f is the frequency of the particle
Since the frequency and wavelength of light are related by the speed of light, c, as:
c = λ * f
We can rearrange the equation to solve for f:
f = c / λ
Substituting this value for f in the equation for velocity:
v = λ * (c / λ)
v = c
Now we can substitute the value of velocity (v) in the equation for momentum:
p = m * c
Now to find the maximum wavelength capable of ionizing sulfur, we need to consider the 1st ionization energy. When the energy of a photon is equal to the ionization energy, the momentum of the electron is given by:
p = √(2 * m * E)
Where:
E is the energy of the photoelectron
We can relate the energy E to the wavelength λ using the equation:
E = h * c / λ
Now, we can substitute the expression for momentum p into the equation for energy E:
√(2 * m * E) = m * c
Simplifying the equation:
2 * m * E = m^2 * c^2
Dividing both sides of the equation by m:
2 * E = m * c^2
Substituting the equation for energy E:
2 * (h * c / λ) = m * c^2
Solving for λ, we get:
λ = 2 * h * c / (m * c^2)
Now, we need the value of the mass of sulfur (m). The atomic mass of sulfur is approximately 32 grams/mole. Converting this to kilograms:
m = 32 / (6.022 x 10^23) kg
Plugging the values of h, c, and m into the equation for λ, we get:
λ = (2 * 6.626 x 10^-34 J·s * 2.998 x 10^8 m/s) / (32 / (6.022 x 10^23) kg * (2.998 x 10^8 m/s)^2)
Simplifying the equation will give us the maximum wavelength of radiation capable of ionizing sulfur and producing the given effect.