Photon energy and momentum

Title

Photon Energy and Momentum

Problem Statement

Light is a particle, and each particle is called a photon. The energy $E$ and momentum $p$ of a single photon corresponding to light with wavelength $\lambda$ are expressed as follows using Planck's constant $h$ :

E = \frac{hc}{\lambda}

p = \frac{h}{\lambda}

Here, $c$ is the speed of light in a vacuum.

Monochromatic light with wavelength $\lambda$ is emitted from a light source. Find the ratio $\frac{E}{p}$ of the energy $E$ and momentum $p$ of a single photon corresponding to this light.

Constraints

$c = 3.0 \times 10^8 \ \text{m/s}$
$h = 6.6 \times 10^{-34} \ \text{J}\cdot\text{s}$
$\lambda = 6.0 \times 10^{-7} \ \text{m}$

Input Format

Enter the value of the calculated ratio $\frac{E}{p}$ in units of $\text{m/s}$ .

Explanation

Properties of Photons and Derivation of the Relationship

The energy $E$ and momentum $p$ of a single photon, as given in the problem statement, can be expressed as follows using the wavelength of light $\lambda$ , Planck's constant $h$ , and the speed of light in a vacuum $c$ .

E = \frac{hc}{\lambda}

p = \frac{h}{\lambda}

Ratio Calculation

Substitute the respective equations into the desired ratio $\frac{E}{p}$ .

\frac{E}{p} = \frac{\frac{hc}{\lambda}}{\frac{h}{\lambda}}

Canceling the common $\frac{h}{\lambda}$ in the numerator and denominator, we get the following:

\frac{E}{p} = c

Therefore, the ratio of a photon's energy to momentum does not depend on the wavelength of light $\lambda$ or Planck's constant $h$ , but is equal to the speed of light in a vacuum $c$ itself.

Numerical Substitution

From the constraints, the speed of light in a vacuum is $c = 3.0 \times 10^8 \ \text{m/s}$ .

Therefore, the value of the ratio we are looking for is $3.0 \times 10^8 \ \text{m/s}$ .

The natural number to enter is 300000000.