# Photons and the Photoelectric Effect

• frankene
In summary, the conversation discusses the use of the equation hf = KEmax + Work Function to determine the wavelength of light incident on the surface of metallic sodium, with a work function of 2.3 eV. The maximum speed of the emitted photoelectrons is 1.26e6 m/s. After converting the work function to joules and using the equation E = hc/lambda, the correct wavelength of the light is determined to be 1.82e-7 m. The conversation also mentions the use of the equation 0.5 * mv^2 to find the kinetic energy of the electron, which is then added to the work function to calculate the energy of the incident light.
frankene
Light is incident on the surface of metallic sodium, whose work function is 2.3 eV. The maximum speed of the photoelectrons emitted by the surface is 1.26e6 m/s. What is the wavelength of the light?

I first converted Work Function to Joules:

2.3eV x (1.6e-19 J / 1eV) = 3.68e-19 J

The equation I know is:

hf = KEmax + Work Function

hf can also be written as:

hc/lamda

I thought I would substitute c w/ the 1.26e6 m/s and use KEmax as zero. My answer was 2.27e-9m and that is not correct.

I used the following:

f = Work Function / h = 5.55e14 Hz

Then I used:

lamda = v/f = 2.27e-9 m

If I can't substitute v for c, then what am I looking for with v? I have looked at this problem for 3 days w/ no new ideas. Anyone else have any?

Thank you!

Have you worked out the ke of the electrons?

When I do this I get the E for the light different to the value you give.

And I don't understand what you're doing with this
lamda = v/f = 2.27e-9 m

Find the Ke of the electron (joules) Convert WF to joules and add. This is the energy of the light.

Then use E = hc/lamda

I get 636 nm

Thank you.

I didn't realize I could find KE by using .5 *mv^2 and use the mass of an electron. Once I did that I added WF and then divied hc by my answer. The answer I got was 1.82e-7 m and that was correct.

Thank you again for the help. I knew I was thinkging too hard and not looking at the obvious.

good - don't know why my answer's wrong...one too many glases of wine before using a calculator?

## 1. What is a photon?

A photon is a fundamental particle of light that carries energy and travels at the speed of light.

## 2. How does the photoelectric effect work?

The photoelectric effect is the process by which photons of light strike a material and transfer their energy to electrons, causing them to be ejected from the material.

## 3. What is the relationship between the frequency of light and the energy of a photon?

The energy of a photon is directly proportional to its frequency. This means that higher frequency light (such as ultraviolet or gamma rays) has more energy per photon than lower frequency light (such as radio waves).

## 4. Can the photoelectric effect be explained by classical physics?

No, the photoelectric effect cannot be fully explained by classical physics. Classical physics predicts that increasing the intensity of light should increase the number of electrons ejected, but in reality, it is the frequency of the light that determines the energy of the ejected electrons.

## 5. How is the photoelectric effect used in everyday life?

The photoelectric effect is used in a variety of modern technologies, such as solar panels, photodiodes, and digital cameras. It is also the basis for devices like photocopiers and barcode scanners.

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