# How many photons are produced in a laser pulse of 0.210 J at 535 nm?

Nov 22, 2016

$5.65 \cdot {10}^{17} \text{photons}$

#### Explanation:

Your strategy here will be to use the Planck - Einstein relation to calculate the energy of a single photon of wavelength $\text{535 nm}$, then use the total energy of the laser pulse to figure out exactly how many photons were needed to produce that output.

The Planck - Einstein relation looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{E = h \cdot \nu}}}$

Here

• $E$ - the energy of the photon
• $h$ - Planck's constant, equal to $6.626 \cdot {10}^{- 34} \text{J s}$
• $\nu$ - the frequency of the photon

Notice that this equation relates the energy of the photon to its frequency. More specifically, it shows that the energy of the photon is directly proportional to its frequency.

In simple terms, the higher the frequency, the more energetic the photon.

Now, the frequency of the photon is related to its wavelength by the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{l a m \mathrm{da} \cdot \nu = c}}}$

Here

• $l a m \mathrm{da}$ - the wavelength of the photon
• $c$ - the speed of light in a vacuum, usually given as $3 \cdot {10}^{8} {\text{m s}}^{- 1}$

Rearrange to solve for $\nu$

$l a m \mathrm{da} \cdot \nu = c \implies \nu = \frac{c}{l a m \mathrm{da}}$

Plug in your value to find -- do not forget to convert the wavelength from nanometers to meters

nu = (3 * 10^8color(red)(cancel(color(black)("m")))"s"^(-1))/(535 * 10^(-9)color(red)(cancel(color(black)("m")))) = 5.6075 * 10^(14)"s"^(-1)

Plug this value into the Planck - Einstein equation to find the energy of a single photon

E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 5.6075 * 10^(14)color(red)(cancel(color(black)("s"^(-1))))

$E = 3.716 \cdot {10}^{- 19} \text{J}$

Now use the total energy of the laser pulse to find how many photons were needed for this output

$0.210 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J"))) * "1 photon"/(3.716 * 10^(-19)color(red)(cancel(color(black)("J")))) = color(darkgreen)(ul(color(black)(5.65 * 10^(17)"photons}}}}$

The answer is rounded to three sig figs.