# Photoelectric effect/finding kinetic energy.

• xxabr
In summary, the photoelectric effect experiment in question uses a light of frequency 8x1014Hz and a metal with a work function of 1.2eV. By using the equation Ek= hf-W, where h=6.63x10-34Js, f=8x1014Hz, and W=1.2eV, the maximum kinetic energy of the emitted electrons is calculated to be 51.84 J. It is important to note that the units must be consistent throughout the calculation, and the stop potential of 1.5V can be considered as adding to the work function, resulting in a decrease in the kinetic energy of the electrons.
xxabr

## Homework Statement

A photoelectric effect experiment uses a light of frequency 8x1014Hz and a metal with a work function of 1.2eV. A 1.5V stop potential is also applied. What is the maximum kinetic energy of the emmited electrons?

## Homework Equations

I was thinking about using Ek= hf-W

## The Attempt at a Solution

h= 6.63x10-14
W= 1.2eV
f= 8x1014

If I were to use that equation I would get:
Ek=hf-W
= (6.63x10-14)(8x1014) - (1.2)
= 53.04 - 1.2
= 51.84 J

* I have never used eV before so I'm not sure if I had to do anything with it...
Also it gives me the stop potential and I'm not sure if or how I'm supposed to use it.

xxabr said:

## Homework Statement

A photoelectric effect experiment uses a light of frequency 8x1014Hz and a metal with a work function of 1.2eV. A 1.5V stop potential is also applied. What is the maximum kinetic energy of the emmited electrons?

## Homework Equations

I was thinking about using Ek= hf-W

## The Attempt at a Solution

h= 6.63x10-14
Look up h again. What number should go in the exponent, 10?, and what are the units that go with h?
W= 1.2eV
f= 8x1014

If I were to use that equation I would get:
Ek=hf-W
= (6.63x10-14)(8x1014) - (1.2)
= 53.04 - 1.2
Watch the units. What units go with "53.04"? What units go with "1.2"?
= 51.84 J

* I have never used eV before so I'm not sure if I had to do anything with it...
It has a lot to do with it. You need to decide what energy units you are going to use -- either Joules or eV -- and convert as necessary.

Also it gives me the stop potential and I'm not sure if or how I'm supposed to use it.
You can think of the energy associated with the stop potential as adding to the work function, reducing the kinetic energy of the electron.

Oops yeah, I made a mistake. h is 6.63x10-34Js.

And I got it.

Thank yoou

Last edited:

## 1. What is the photoelectric effect?

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

## 2. How does the photoelectric effect work?

The photoelectric effect occurs when a photon of light strikes the surface of a material and transfers its energy to an electron, causing it to be ejected from the material. The minimum energy required for this to happen is known as the material's work function.

## 3. What is the equation for finding the kinetic energy of an electron emitted by the photoelectric effect?

The equation is KE = hf - Φ, where KE is the kinetic energy of the electron, h is Planck's constant, f is the frequency of the incident light, and Φ is the work function of the material.

## 4. What factors affect the kinetic energy of the emitted electrons in the photoelectric effect?

The kinetic energy of the emitted electrons depends on the frequency of the incident light, the work function of the material, and the intensity of the light. Increasing the frequency or intensity of the light will result in higher kinetic energies of the emitted electrons.

## 5. How is the photoelectric effect used in practical applications?

The photoelectric effect is used in many practical applications including photovoltaic cells (solar panels), photocells (used in light meters and burglar alarms), and photomultiplier tubes (used in particle detectors). It also played a crucial role in the development of quantum mechanics and our understanding of the behavior of light and matter.

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