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Wave Optics

Question
CBSEENPH12040098

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

  • 0.33 space cross times space 10 to the power of 6
  • 7 cross times 10 to the power of negative 24 end exponent
  • 10 to the power of negative 22 end exponent
  • 5 cross times 10 to the power of negative 22 end exponent

Solution

D.

5 cross times 10 to the power of negative 22 end exponent

Energy of photon is given by
                 straight E space equals space hc over straight lambda                                 ...(i)
Where h is Planck's constant, c the velocity of light and straight lambda its wavelength.
de-Broglie wavelength is given by
                         straight lambda space equals space straight h over straight p                           ...(ii)
           p being momentum of photon. 
From Eqs. (i) and (ii), we can have         
                            straight E space equals space fraction numerator hc over denominator straight h divided by straight p end fraction space equals space pc
               or space space space space space space space space space space space space space space space straight p space equals space straight E divided by straight c
Given comma space space straight E space equals space 1 space MeV space equals space 1 cross times 10 to the power of 6 cross times 1.6 cross times 10 to the power of negative 19 end exponent straight J
                straight C space equals 3 space cross times space 10 to the power of 8 space straight m divided by straight s
Hence, after putting numerical values, we obtain
            straight p equals fraction numerator 1 cross times 10 to the power of 6 cross times 1.6 cross times 10 to the power of negative 19 end exponent over denominator 3 cross times 10 to the power of 8 end fraction kgm divided by straight s
space space space equals space 5 space cross times space 10 to the power of negative 22 end exponent space kgm divided by straight s

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