# Classical physics wavelength of monochromatic light

• Stevee
In summary, the stopping potential of 0.50 V is required when a phototube is illuminated with monochromatic light of wavelength 600 nm. Monochromatic light of a different wavelength is now shone on the tube, and the stopping potential is measured to be 1.1V. The wavelength of this new light is 113x10^-8 m.
Stevee

## Homework Statement

A stopping potential of 0.50 V is required when a phototube is illuminated with monochromatic light of wavelength 600 nm. Monochromatic light of a different wavelength is now shone on the tube, and the stopping potential is measured to be 1.1V. What is the wavelength of this new light? (c = 3.00 × 108 m/s, e = - 1.60 × 10-19 C, h = 6.626 × 10-34 J · s, 1 eV = 1.60 × 10-19 J)

## Homework Equations

E=h*f (f=c/lambda)
h*f=sigma+Ek

## The Attempt at a Solution

E=h*c/lambda
lambda=h*c/E

I am just unsure on how to obtain the energy for the new monochromatic light given another.
Thankyou!

How do you incorporate the stopping potential in the equation?
What does it tell you about the problem?

So, the stopping potential would be equivalent to the available kinetic energy which is [Avail. Ek=E - sigma] right? Do i then work out the work function from the other provided wavelength?
The stopping potential tells us the amount of voltage required to stop the movement of electrons, so into the tube?

The stopping potential is used to bring the emitted photoelectrons to a halt.
How can you calculate the kinetic energy of the photoelectrons from this?

Oh, sorry photoelectron not electron.. my mistake
well I am not entirely sure, I am having a real blank at the moment... I am not sure on whether its possible to use the first wavelength (and stopping pot.) to calculate the work function to sub into the equation, but the frequency changes with wavelength and therefore so does the kinetic energy, maybe I am overthinking this?

What is the relationship between electric charge and voltage?

q=E/V?
oh woww.. haha
You use E=qV to find the energy then just sub it into E=h*c/lambda and solve for lambda right?..

Yes, the electric potential difference or voltage between two points tells us the amount
of energy expended by or given to each coulomb of charge moving between the two points.

Sorry, I meant to say that the calculation gives you the kinetic energy of the
emitted photoelectrons, not the energy of the photons, but your equation is correct.

Thankyou,
hmm it doesn't seem to be working out for me.
So i used E=q*V
E=(1.6x10^-19)*(1.10)=1.76x10^-19
into... E=h*c/lambda
1.76x10^-19=(6.63x10^-34)*(3x10^8)/lambda
where lambda comes out as113x10^-8m.
im doing something wrong

Last edited:
The voltage tells you something about the photoelectrons, not the photons!
Since it is stopping the photoelectrons, what do you think it is a measure of?

so it would be equivalent to the energy required to emit a photoelectron from the tube?

No, the energy you calculated is what a photoelectron loses when it
goes through that potential, so it is its kinetic energy as it leaves the material.

Oh ok, ofcourse haha, so from this we can work out the photon energy?

Set the equations up for the two different photons

hf = ∑ + EK

600nm..
[6.64x10^-34*3x10^8]/600x10^-9 = ∑ + 1.76x10^-19
unknown nm...
[6.64x10^-34*3x10^8]/ lambda = ∑ + 1.76x10^-19

It is easier to see what is going on if you do this

hf1 = ∑ + EK1

hf2 = ∑ + EK2

## What is the definition of "Classical physics wavelength of monochromatic light"?

The classical physics wavelength of monochromatic light refers to the distance between two consecutive peaks or troughs of a wave of single color or frequency.

## How is the classical physics wavelength of monochromatic light calculated?

The classical physics wavelength of monochromatic light can be calculated by dividing the speed of light by the frequency of the light wave. This is represented by the equation λ = c/f, where λ is the wavelength, c is the speed of light, and f is the frequency.

## What is the relationship between wavelength and frequency in classical physics?

In classical physics, the wavelength and frequency of a light wave are inversely proportional. This means that as the wavelength increases, the frequency decreases, and vice versa.

## How does the wavelength of monochromatic light affect its behavior?

The wavelength of monochromatic light affects its behavior in various ways, including its ability to diffract, reflect, and refract. The shorter the wavelength, the more easily the light can diffract and reflect, while longer wavelengths are better at refracting.

## What are some real-world applications of understanding the classical physics wavelength of monochromatic light?

Understanding the classical physics wavelength of monochromatic light is crucial in various fields, such as optics, astronomy, and telecommunications. It allows us to design and develop technologies such as telescopes, lasers, fiber optics, and communication devices.

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