# the number of photons emitted per second by the laser? (Plank’s constant Light of wave length 6730 Å is emitted by a He-Ne laser with output power of 2.3 mW. Calculate h = 6.63 × [10]^(-34) J)

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May 30, 2016

The intensity or power output of a laser ($P$) is equal to $n h f$, where $n$ is the number of photons emitted per second, $h$ is Planck's constant and $f$ is the frequency.

Therefore, by rearranging:
$n = \frac{P}{h f}$

Since we don't have frequency but rather wavelength, we can use $c = f \lambda$ (rearranged to give $f = \frac{c}{\lambda}$) to give a final, more useful equation:
$n = \frac{P}{h \frac{c}{\lambda}}$

One further rearrangement, though you could use the above form:
$n = \frac{P \lambda}{h c}$

We need to get everything into the correct units before we apply the equation.
The power output is $2.3 m W$, this must be in Watts (so $2.3 \cdot {10}^{-} 3$ Watts).
The wavelength must be in metres not Ångströms and since $1$ Ångström $= {10}^{-} 10$ metres, then our wavelength in metres is $6370 \cdot {10}^{-} 10$.
$c$ represents the speed of light $3 \cdot {10}^{8} m {s}^{-} 1$
$h$ is Planck's constant - $6.63 \cdot {10}^{-} 34 J s$

So finally, $n = \frac{2.3 \cdot {10}^{-} 3 \cdot 6370 \cdot {10}^{-} 10}{6.63 \cdot {10}^{-} 34 \cdot 3 \cdot {10}^{8}}$
$n = 7.37 \cdot {10}^{15} {s}^{-} 1$

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