# A 600W mercury lamp emits monochromatic radiation of wavelength 331.3 nm. How many photons are emitted from the lamp per second?

May 10, 2017

$\frac{1.00 \cdot {10}^{21} \text{ photons}}{s}$

#### Explanation:

Let's first write down our givens for this problem

Given
$\textcolor{g r e e n}{\text{Power} = 600 W \mathmr{and} 600 \frac{J}{s}}$
color(green)(lambda = 331.3" nm" or (3.313*10^-7" m"))

Now let's try to establish a few things.

The mercury lamp is emitting a certain amount of $\text{Energy}$ per $\text{second}$, defined as its $\text{Power}$. The source is coming from a monochromatic radiation with a wavelength of $3.313 \cdot {10}^{-} 7 \text{ m}$. We are asked to find how many photons are being emitted from the lamp that is providing $600 \text{ J}$ of energy per second.

$- - - - - - - - - - - - - - - - - - - - -$

$\textcolor{b l u e}{\text{Step 1: Figure out the energy associated with the photon}}$

We use the following formula

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a} \textcolor{m a \ge n t a}{E = h \cdot f}$

Where
$\text{E = Energy of the photon (J)}$
"h = Planck's constant" (6.62*10^-34" J*s")
$\text{f = frequency of the photon} \left(\frac{1}{s}\right)$

But we aren't given the frequency; we are give the wavelength.

Well, we know that the speed of light is constant and given as $3.00 \cdot {10}^{8} \frac{m}{s}$ and can be calculated by using the following:

$\textcolor{w h i t e}{a a a a a a a a a a a a a} \textcolor{m a \ge n t a}{c = f \cdot \lambda}$

Where
$c = \text{speed of light} \left(3.00 \cdot {10}^{8} \frac{m}{s}\right)$
$f = \text{frequency of radiation} \left(\frac{1}{s}\right)$
$\lambda = \text{wavelength} \left(m\right)$

Knowing this, we can isolate to solve for the frequency

$\textcolor{w h i t e}{a a a a a a a a a a a a a} \textcolor{m a \ge n t a}{c = f \cdot \lambda} \to \frac{c}{\lambda} = f$

We can replace $\text{f}$ in the energy of the photon equation with $\frac{c}{\lambda}$ and solve.

$\textcolor{m a \ge n t a}{E = h \cdot f} \to E = \frac{h c}{\lambda}$

• E =[(6.62*10^-34 J*cancel"s")(3.00*10^8cancel"m"/cancel"s")]/(3.313*10^-7 cancel"m")->color(orange)(5.99 * 10^-19" J")

$\textcolor{b l u e}{\text{Step 2: Use dimensional analysis to figure out photons emitted per s}}$
Well, what does this tell us? This tells us that 1 photon has $5.99 \cdot {10}^{-} 19 \text{ J}$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a} \left(1 \text{ photon")/(5.99*10^-19" J}\right)$

We were also told that the lamp produced $600 \text{ J}$ per second

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a} \frac{600 \text{ J}}{s}$

We can use dimensional analysis to figure out the number of photons emitted per second.

color(white)(aaaaa)(600 cancel"J")/s*(1" photon")/(5.99*10^-19 cancel"J")->color(orange)[(1.00*10^21" photons")/s]

$\frac{\textcolor{\mathmr{and} a n \ge}{\text{Answer": (1.00*10^21" photons}}}{s}$