Photoelectric Question

Photoelectric effect Question?

Whether we choose photon model or wave model, we know that we need to supply some energy for the electrons to come out of the metal.
Instead of having frequency more than cut off, suppose we shine lots of light of lower frequency for a long time, then also we are giving lots of energy. So the electrons should come out. But they do not. This cannot be explained by wave model.

Photon model explains this by saying that light consists of photons. Each photon has energy hf, where h = Planck's constant, f = frequency

When photons strike a metal surface, then an electron in the metal absorbs a photon completely or does not absorb at all. When we shine high intensity light of lower frequency, then what it means is that lots of photons are hitting the metal surface. When an electron absorbs any such photon, then it does not get enough energy to come out.

Photoelectric effect question?

I need help with the following question, not really sure how to do it. The question is;
In a photoelectric effect experiment, it is found that a stopping potential of 1 V is needed to stop electrons when light of wavelength 2600 Angstrom is used, and 2.3 v when light of wavelength 2070 Angstom is used. Determine Planck's constant (h) and the work function of the emitter.

The relevant equations are, I believe, Kmax = hf - Work function
and Kmax= -eV(stopping), My attempt at calculating h resulted in something that was quite a lot off the actual value of 6.626 x 10^-34, so any help is much appreciated.

Physics (Photoelectric Question)?

You want to use a material that has a work function that has an energy equal to the energy of a photon of visible light.

The energy of a photon as a function of wavelength is given by:

E(L) = h*c/L

where L is the wavelength, h is Planck's constant = 6.6261 * 10^-34 m^2*kg / s, and c is the speed of light in a vacuum = 2.998 * 10^8 m/s

Visible light has wavelengths between about 400 nm and 700 nm (1 nm = 10^-9 m).

Rearranging the above equation to solve for the wavelength in terms of energy, we get that:

L(E) = h*c/E

In order to make the units come out right, we need to convert the energies you are given in eV into joules. 1 electron volt (eV) = 1.6022 * 10^-19 joules.

Using these relationships, one calculates that:

A 4.2 eV photon has a wavelength of 276 nm
A 4.5 eV photon has a wavelength of 295 nm
A 2.5eV photon has a wavelength of 496 nm
A 2.3 eV photon has a wavelength of 539 nm

The 2.5 eV and 2.3eV cases (Ba and Li, respectively) have energies that correspond to the energies of photons of visible light.

Physics question on photoelectric effect?

Image intensifiers used in night-vision devices create a bright image from dim light by letting the light first fall on a photocathode. Electrons emitted by the photoelectric effect are accelerated and then strike a phosphorescent screen, causing it to glow more brightly than the original scene. Recent devices are sensitive to wavelengths as long as 900 {\rm nm}, in the infrared:

A) If the threshold wavelength is 900 {\rm nm}, what is the work function of the photocathode?

B) If light of wavelength 500 nm strikes such a photocathode, what will be the maximum kinetic energy, in {\rm eV}, of the emitted electrons?

Photoelectric Effect question help please?

I'll assume that wavelength is in meters.

Each photon of light brings an energy

E = hf = hc/λ

to an electron. Some of that energy will be used to break the electron free of the material (that's the work function, φ) and the remainder will appear as kinetic energy of the electron. So

E = K + φ

That's not the way you usually see the equation for the photoelectric effect, but that's what the usual equation really means.

So first, let's find the energy of one of the incident photons

E = hc/λ = (4.136×10^-15 eV∙s)(2.998×10^8 m/s) / (3.0×10^-7 m) = 4.13 eV

where I've used h in eV∙s to save on unit conversion.

You'll note that 4.13 eV is less than the energy required to break an electron free of mercury, so of those three, only lithium and iron will exhibit the photoelectric effect.

Now we can solve the photoelectric effect equation for K

K = E - φ

For lithium

K = 4.13 eV - 2.3 eV = 1.83 eV

and for iron

K = 4.13 eV - 3.9 eV = 0.23 eV

The difference between your answers and mine is probably the number of digits used for h and c.

Conceptual photoelectric effect question?

The photoelectric effect is where light knocks electons loose from a metal surface. It takes a light of at least a certain frequency to knock loose an electron. Light with a lower frequency (longer wavelength), no matter how bright, does not knock loose electrons. Why does this evidence support the idea of photons? Explain using the concept of energy. Explain why the wave theory of light fails to explain the photoelectric effect.

Thanks