Question

The average intensity of an electromagnetic wave is $\mathit{\lambda }$, where E₀ is the amplitude of the electric field portion of the wave. Find a general expression for the photon flux in terms of E₀ and wavelength $\mathit{\lambda }$.

Short answer

The general expression for the photons flux in terms of $\lambda$ and E₀ will be given as$j=\frac{{\epsilon }_0\lambda E_0^2}{2h}$

Step-by-step solution

Step-1: Calculation of average intensity of an electromagnetic wave.

To find an expression for the photon flux j (number of photons/s.m²), the average intensity of an electromagnetic wave will be required,

$\mathit{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}\mathit{c}{\mathit{E}}_{\mathbf{0}}^{\mathbf{2}}$……………..(1)

Here E₀ electric field intensity, I is the intensity of light, c is the speed of light in the vacuum, and ${\mathit{\epsilon }}_{\mathbf{0}}$is the permittivity of free space

The intensity of light can be written as:

$\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}$………........……..(2)

The general equation for power,

$\mathit{P}\mathbf{=}\frac{\mathbf{E}}{\mathbf{t}}$……………………(2)

Here, P is the power, E is the energy and t is the time.

Energy E of a photon of wavelength $\mathit{\lambda }$can be written as:-

$\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda }}$

Here, h is the Plank’s constant, and c is the speed of light in the vacuum.

Step-2: Derivation of the final formula

Substitute value from equation (2) into equation (1), and we get,

$\frac{P}{A}=\frac{1}{2}{\epsilon }_{0}C{E}_0^2$

Substitute value from equation (3) into the above expression and we get,

$\begin{array}{rcl}\frac{\left({\displaystyle \frac{E}{t}}\right)}{A}& =& \frac{1}{2}{\epsilon }_{0}C{E}_0^2\\ \frac{E}{At}& =& \frac{1}{2}{\epsilon }_{0}C{E}_0^2\end{array}$

Here, E is the energy of the photons.

Step-3: Final Formula

The total energy will be n times the energy of each photon, where n is the number of photons. So the expression for total energy E_n is given by:-

$\begin{array}{rcl}{E}_n& =& n\left(\frac{hc}{\lambda }\right)\\ & =& \frac{nhc}{\lambda }\end{array}$

Expression for total energy:

$\frac{E_n}{At}=\frac{1}{2}{\epsilon }_{0}C{E}_0^2$

Substituting $E_n=\frac{nhc}{\lambda }$in the above expression and we get

$\begin{array}{rcl}\frac{1}{At}\left(\frac{nhc}{\lambda }\right)& =& \frac{1}{2}{\epsilon }_{0}c{E}_0^2\\ \left(\frac{nhc}{\lambda }\right)& =& \frac{1}{2}{\epsilon }_{0}c{E}_0^2\\ \frac{nh}{At\lambda }& =& \frac{1}{2}{\epsilon }_0{E}_0^2\\ \frac{n}{At}& =& \frac{{\epsilon }_0\lambda E_0^2}{2h}\end{array}$

Therefore, the expression for the photon flux j is given by $j==\frac{{\epsilon }_0\lambda E_0^2}{2h}$.