17.5.2 Momentum of Photon

In two papers published in 1909 and 1916, Einstein pointed out that for photons to be full-fledged particles, they must also carry momentum

\displaystyle p=\frac{h}{\lambda }

But how can photons, which have no mass, have momentum? Well well well. In 1922, the American physicist Compton directed a stream of x-ray photons at carbon atoms. If the photons do have momentum, then their momentum should decrease after colliding with the electrons (initially bound) in the carbon atoms. This was indeed confirmed by the observation that the wavelengths of the x-ray increased after being scattered (by the carbon atoms). Furthermore, calculations based on PCOM and PCOE confirmed that the numbers checked out. So \displaystyle p=\frac{h}{\lambda } was validated.

Example

A laser beam is incident normally on a mirror. The beam has wavelength 680 nm and power 14 mW. Calculate the force exerted by the laser on the mirror if it is 100% reflective.

Solution

Momentum of each laser photon, \displaystyle {{p}_{{photon}}}=\frac{h}{\lambda }=\frac{{6.63\times {{{10}}^{{-34}}}}}{{680\times {{{10}}^{{-9}}}}}=9.75\times {{10}^{{-28}}}\text{ kg m }{{\text{s}}^{{-1}}}

Energy of each laser photon, \displaystyle {{E}_{{photon}}}=\frac{{(6.63\times {{{10}}^{{-34}}})(3.00\times {{{10}}^{8}})}}{{680\times {{{10}}^{{-9}}}}}=2.925\times {{10}^{{-19}}}\text{ J}

Number of photons incident on mirror per unit time, \displaystyle {{N}_{t}}=\frac{P}{{{{E}_{{photon}}}}}=\frac{{0.014}}{{2.925\times {{{10}}^{{-19}}}}}=4.786\times {{10}^{{16}}}\text{ }{{\text{s}}^{{-1}}}

Force exerted \displaystyle =\frac{{dp}}{{dt}}={{N}_{t}}\times \Delta p=(4.786\times {{10}^{{16}}})(2\times 9.75\times {{10}^{{-28}}})=9.33\times {{10}^{{-11}}}\text{ N}

Concept Test

3449

Interesting

Crooke’s Radiometer

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