# De Broglie wavelength of a photon

• StudentofPhysics
In summary, the wavelength of a 5.0 eV photon is 9.82 x 10^-18 meters. The de Broglie wavelength of a 5.0 eV electron is 3.41e-20 meters.
StudentofPhysics
1. What is the wavelength of a 5.0 eV photon?
What is the de Broglie wavelength of a 5.0 eV electron?

2.E = hc/lambda
lambda = h/momentum

3. I know the first one simply 248 nm for the wavelngth.

I don't know how to find the de Broglie though since I don't know what the momentum of the electron is.

Try finding it form:
$$E=\frac{p^2}{2m}$$

ok i tried it with that formula this way:

5 eV = p^2 / (9.11 x 10^-31kg)
p= 6.75 x 10^-17

this gave me a wavelength of 9.82 x 10^-18 m and wasn't correct...

I'm still doing something wrong

I'm having the same problem with this question:
An electron, starting from rest, accelerates through a potential difference of 415 V. What is the final de Broglie wavelength of the electron, assuming that its final speed is much less than the speed of light?

Applying that formula i get 3.41e-20

For both of these questions I figured maybe it was in nm, so i switched to meters by adding ^-9 and they were still wrong...

StudentofPhysics said:
ok i tried it with that formula this way:

5 eV = p^2 / (9.11 x 10^-31kg)
p= 6.75 x 10^-17

this gave me a wavelength of 9.82 x 10^-18 m and wasn't correct...I'm still doing something wrong

You probably didn't converted energy from eV in Joules. And the formula is:

$$E=\frac{p^2}{2m_{e}}$$

(2 is missing in yours). I got the result $\lambda=5.48\cdot{10^{-10}}m$.

Last edited:
the 2 and converting to joules were the problems.

Much appreciated, I've got it now.

what would i do for the problem with E=415V since I can't just convert that into Joules?

The potential difference is 415v. This is not an energy. Recall the definition of the electron-volt...

It would be very useful if you studied theory once again before trying to solve any more problems, becoase if you have problems on this level it means that you haven't quite got the grasp of meaning of things and you will not be able to make any progress.

thank you for the tips, I figured it out on reviewing the theory. Misunderstood the 415V, its been awhile since I worked with them

## What is the De Broglie wavelength of a photon?

The De Broglie wavelength of a photon is a concept in quantum mechanics that describes the wavelength of a photon, which is a particle of light. It is named after French physicist Louis de Broglie, who first proposed the idea that particles can also exhibit wave-like behavior.

## How is the De Broglie wavelength of a photon calculated?

The De Broglie wavelength of a photon can be calculated using the equation: λ = h/p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the photon. This equation shows the inverse relationship between the momentum and wavelength of a photon, meaning that as the momentum increases, the wavelength decreases.

## What is the significance of the De Broglie wavelength of a photon?

The De Broglie wavelength of a photon is significant because it demonstrates the wave-particle duality of light. This means that light, which is traditionally thought of as a wave, can also behave like a particle. It also helps to explain phenomena such as diffraction and interference in light.

## How does the De Broglie wavelength of a photon relate to the energy of the photon?

The De Broglie wavelength of a photon is inversely proportional to its energy. This means that as the wavelength decreases, the energy of the photon increases. This relationship is important in understanding the behavior of photons in quantum mechanics and in various applications such as laser technology.

## Can the De Broglie wavelength of a photon be observed experimentally?

Yes, the De Broglie wavelength of a photon can be observed experimentally using methods such as the double-slit experiment. This experiment demonstrates the wave-like behavior of photons as they pass through two slits and interfere with each other, creating a pattern on a screen. The resulting pattern can only be explained by considering the De Broglie wavelength of the photons.

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