Electrons are ejected from a surface with speeds ranging up to 6.0 x 10^5 m/s when light with a wavelength of 400 nm shines on it.
a.) What is the work function of this surface?
b.) What is the threshold frequency for this surface?
c.) What is the deBroglie wavelength of an electrion traveling at its maximum velocity?
Please do not just retype quesations and ask us to do them for you. Read your assignment and look for the equation describing the photoelectric effect. The work function is the greek leter "phi" . The kinetic energy of the emitted charge is given by
E = h* f - e * phi
h is Planck's constant and f is the frequency and e is in this case the charge of an electron.
(b) The threshold frequency is the minimum that it takes to get any emission of electrons as all. Get it by setting E = 0 in the equation above and solving for f.
(c) Familiarize your self with the relationship between velocity and De Broglie wavelength of a particle:
(wavelength) = h/(velocity)
For the velocity, use 6.0 x 10^5 m/s
Im not retyping the wuestios hoping for someone to do them for me. I'm not suppose to be in this class and I've been lost all year. Now its time to study for my final and I need to do good, so Im hoping that you or anyone else can give me the equatoin to use and point me in the right direction.
<<I'm hoping that you or anyone else can give me the equation to use and point me in the right direction.>>
If you are not supposed to be in the class, why are you taking it? You will learn much more from the class if you make an effort to at least find the correct equation. I will be happy to critique your work when I see some evidence of it.
I apologize if my previous response came across as dismissive. I understand that you need help, and I'm here to assist you.
The photoelectric effect can be described by the equation:
E = hf - φ
Where:
E is the kinetic energy of the emitted electron
h is the Planck's constant (6.626 x 10^(-34) J·s)
f is the frequency of the incident light
φ (phi) is the work function of the surface (the minimum energy needed to remove an electron from the surface)
Now, let's apply this equation to the given information.
(a) To determine the work function (φ), we need to use the equation with the known values of E, h, and f. We need to find the frequency (f) from the given wavelength (λ) of the incident light:
c = λf
Where:
c is the speed of light (approximately 3.00 x 10^8 m/s)
λ is the wavelength of the light (400 nm = 400 x 10^(-9) m)
Rearranging the equation, we get:
f = c / λ
Substituting the values, we have:
f = (3.00 x 10^8 m/s) / (400 x 10^(-9) m)
Now, we can use this value of frequency (f) along with the given kinetic energy (E) to solve for the work function (φ):
E = hf - φ
Note that the kinetic energy is related to the speed (v) of the emitted electron by:
E = (1/2)mv^2
Where:
m is the mass of the electron (9.109 x 10^(-31) kg)
You can rearrange this equation to solve for v:
v = √(2E / m)
Substituting the given kinetic energy (E) and mass (m), you can calculate the speed (v) of the emitted electron.
(b) The threshold frequency (f_threshold) is the minimum frequency required to start the emission of electrons from the surface. In this case, we need to find the frequency (f_threshold) that gives us a kinetic energy (E) of zero.
Setting E = 0 in the equation E = hf - φ, we get:
0 = hf_threshold - φ
Now, you can solve for the threshold frequency (f_threshold) using the known value of the work function (φ).
(c) The de Broglie wavelength (λ_db) of a particle is related to its velocity (v) by the equation:
λ_db = h / mv
You can substitute the given maximum velocity (v) into this equation along with the mass of the electron (m) to find the de Broglie wavelength (λ_db) of the electron.
Remember to use the correct units for all quantities (e.g., joules, meters, seconds) and be mindful of conversion factors if necessary.
I hope this explanation helps. Let me know if you have any further questions.