Question

There is a simple relationship between the energy for a photon of light Ephoton in units of eV (electron volts) and its wavelength λ in nm (nanometers) given by: Ephoton (eV) = (1240 eV nm) / (λ nm). For example, if light has a wavelength of 620 nm, then its energy Ephoton = (1240 eV nm) / (620 nm) = 2 eV. Visible light has photon energies of ~1.7 to 3 eV and wavelengths of ~400 to 750 nm (λBLUE = 475 nm, λGREEN = 510 nm, λYELLOW = 570 nm, and λRED = 650 nm).

What is the wavelength λ in nm for light with a photon energy Ephoton = 2 eV?


What is the photon energy for light with a wavelength λ = 630 nm?

In the electromagnetic spectrum, x-rays have much higher photon energies (~100 to 100,000 eV) and shorter wavelengths (~0.01 to 10 nm) than visible light. What is the photon energy for an x-ray with a wavelength λ = 6 nm?

Answer 1

`\lambda=620\ nm`

`E_p=1.9682\ eV`

`E_p=206.67\ eV` is the photon energy of an x-ray

Step-by-step explanation:

Given:

Energy of photon,
`E_p=2\ eV`

Then according to given:

`E_p=(1240)/(\lambda)`

`2=(1240)/(\lambda)`

`\lambda=620\ nm` is the wavelength

Again,

Energy of photon,
`E_p=?`

wavelength of the light,
`\lambda=630\ nm`

`E_p=(1240)/(630)`

`E_p=1.9682\ eV`

Now, photon energy for an x-ray:

wavelength,
`\lambda=6\ nm`

`E_p=(1240)/(6)`

`E_p=206.67\ eV` is the photon energy of an x-ray