Question

How many photons per second must strike a surface of area \(10.0 \mathrm{~m}^{2}\) to produce a force of \(0.100 \mathrm{~N}\) on the surface, if the wavelength of the photons is 600 . \(\mathrm{nm}\) ? Assume that the photons are absorbed.

Short answer

Answer: Approximately \(9.05 \times 10^{19}\) photons per second must strike the surface to produce a force of \(0.100\) N.

Step-by-step solution

Find the frequency of the photons

We are given the wavelength \(\lambda=600~\mathrm{nm}\), and we need to find the frequency \(f\). We can use the relationship between the speed of light (\(c\)), wavelength, and frequency: $$ c = \lambda \cdot f $$ Solve for \(f\): $$ f = \frac{c}{\lambda} $$ Plug in the given values: $$ f = \frac{3.0 \times 10^8~\mathrm{m/s}}{6.0 \times 10^{-7}~\mathrm{m}} = 5.0 \times 10^{14}~\mathrm{Hz} $$

Calculate the energy of a single photon

We can now use Planck's equation to find the energy \(E\) of a single photon: $$ E = h \cdot f $$ Plug in the calculated frequency and Planck's constant: $$ E = (6.63 \times 10^{-34}~\mathrm{Js}) \cdot (5.0 \times 10^{14}~\mathrm{Hz}) = 3.315 \times 10^{-19}~\mathrm{J} $$

Determine the momentum of a single photon

Now we find the momentum \(p\) of a single photon using the energy and speed of light: $$ p = \frac{E}{c} $$ Plug in the calculated energy and the speed of light: $$ p = \frac{3.315 \times 10^{-19}~\mathrm{J}}{3.0 \times 10^8~\mathrm{m/s}} = 1.105 \times 10^{-27}~\mathrm{kg \cdot m/s} $$

Calculate the number of photons per second

Finally, we need to find the number of photons \(n\) hitting the surface per second in order to produce the given force \(F\). We use the relationship between force, momentum, and the number of photons: $$ F = \frac{dp}{dt} = \frac{d}{dt} \left(\frac{E}{c}n\right) $$ Since we assume that the photons are absorbed, the total change in momentum is given by the momentum of a single photon multiplied by the number of photons per second: $$ F = \left(\frac{E}{c}\right) n $$ Solve for \(n\): $$ n = \frac{F \cdot c}{E} $$ Plug in the known values: $$ n = \frac{0.100~\mathrm{N} \cdot 3.0 \times 10^8~\mathrm{m/s}}{3.315 \times 10^{-19}~\mathrm{J}} = 9.05 \times 10^{19}~\mathrm{photons/s} $$ Thus, approximately \(9.05 \times 10^{19}\) photons per second must strike the surface to produce a force of \(0.100\) N.