Photoelectric effect and a metal plate

In summary, the conversation discusses a question on a physics exam involving the Planck constant and the photoelectric effect. The problem provides information on the energy and wavelength of two different monochromatic lights, and the goal is to determine the Planck constant. The conversation also mentions the concept of work function and provides two equations to solve for the Planck constant and work function.
  • #1
Bindle
13
0
First of all, Hi we haven't even learned the E = ((hc)/λ) in the physics class but we have this question on our test-exam and I hope that it's fairly easy:

Homework Statement



A metal plate is shone upon with a monochromatic light. When the wave-length is 550 nm a photo-electric effect is observed where the fastest photo electrons have the energy of 0.45 eV. When the light have 410 nm wave-length the quickest photo electrons get the energy 1.26 eV. Decide the Planck constant with help of these values.

First λ = 550 x 10^9 m
First E = 0.45 eV
Second λ = 410 x 10^9 m
Second E = 1.26 eV

Homework Equations



E = h*v
λν = c

The Attempt at a Solution



Ok, so I start out with, I'm not sure if I even is to use the conversion of eV to J so I can find the constant but it's the only way I can see that we can do this:

0.45 x (1.60217657 × 10^-19) = E

c/λ= v

(3*10^8)/(550 x 10^9) = v

E/v = h

(0.45 x (1.60217657 × 10^-19))/((3*10^8)/(550 x 10^9)) = h

1.3217957e-16 Js = h

Which doesn't look right. But I guess this has to do with photoelectric effect which haven't learned anything about yet, and that I should use the two values and wavelengths in co-operation in some form to get the Plancks constant.

Anybody who could make me see what to do?

Bindle
 
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  • #2
You need an additional parameter, to wit, the work function of the metal.

Basically, E = hf - W
where W is the work function of the metal, f is the frequency of the impinging light, and E is the maximum observed kinetic energy of the emitted electrons. E cannot be negative.

So you have 2 equations and 2 unknowns: h and W. Solve for h.
 

1. What is the photoelectric effect?

The photoelectric effect is the phenomenon where electrons are emitted from a material when it is exposed to light of a certain frequency. This effect was first observed by Heinrich Hertz in 1887 and later explained by Albert Einstein in 1905.

2. How does the photoelectric effect work?

The photoelectric effect occurs when photons (particles of light) strike the surface of a material, transferring their energy to electrons within the material. If the energy of the photons is high enough, they can knock electrons out of the material, resulting in the emission of electrons.

3. What is a metal plate?

A metal plate is a flat piece of metal that is usually made of a single type of metal, such as aluminum, copper, or gold. It is often used as a conducting surface in experiments involving the photoelectric effect.

4. How does a metal plate affect the photoelectric effect?

A metal plate is necessary for the photoelectric effect to occur because it provides a surface for the photons to strike and a source of free electrons for them to interact with. The type of metal used can also affect the energy of the electrons emitted.

5. What are the practical applications of the photoelectric effect?

The photoelectric effect has many practical applications, including in solar panels, photocells, and photomultiplier tubes. It is also used in devices such as photocopiers, digital cameras, and smoke detectors.

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