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

Answer:

$\lambda=620\ nm$

$E_p=1.9682\ eV$

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

Explanation:

Given:

Energy of photon, $E_p=2\ eV$

Then according to given:

$E_p=\frac{1240}{\lambda}$

$2=\frac{1240}{\lambda}$

$\lambda=620\ nm$ is the wavelength

Again,

Energy of photon, $E_p=?$

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

$E_p=\frac{1240}{630}$

$E_p=1.9682\ eV$

Now, photon energy for an x-ray:

wavelength, $\lambda=6\ nm$

$E_p=\frac{1240}{6}$

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