The momentum of a photon of energy $$1$$ MeV in kg $$m/s$$ will be

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Infinity Learn · Accepted Answer

Energy of photon is given $$byE=hc$$λ   .    .   .   iWhere  h  is  plank's  constant,  c  the  velocity  of  light  and  λ   its  wavelength.de−Broglie  wavelength  is given  byλ$$=hp$$   .     .     .    iiP  being  momentum  of  photon  from Eq.s  i  and  ii,  we  can  $$haveE=hch/p=pc$$   or   $$p=E/cGiven$$,  $$E=1$$  $$MeV=1$$×$$106$$×$$1.6$$×$$10$$−$$19Jc=3$$×$$108m/sHence$$,   after  putting   numerical  values,   we  obtain   $$p=1$$×$$106$$×$$1.6$$×$$10$$−$$193$$×$$108kgm/s=5$$×$$10$$−$$22kgm/s$$

The momentum of a photon of energy 1 MeV in kg m/s will be

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