Question

The following three particles all have the same total energy E: (a) a photon, (b) a proton, and (c) an electron. Rank the magnitudes of the particles’ momenta from greatest to smallest.

Answer 1

Answer:

The required order will be  p₁  > p₂  > p ₃ .

Explanation:

Momentum of a particle is given by :-

p = $\sqrt{2mE}$

So, Momentum of a proton will be :-

p₂ = $\sqrt{2m_{2} E}$

And Momentum of an electron will be :-

p₃ = $\sqrt{2m_{3} E}$

where m₂ = mass of a proton = 1.67 × 10⁻²⁷ kg

            m₃ = mass of an electron = 9.1 × 10⁻³¹ kg

Also, Momentum of a photon is given by :-

p₁ =   $\dfrac{E}{c}$

     where c = speed of light in vacuum = 3 × 10⁸ m/s.

Thus, from equations (1), (2) , (3), we get

         p₁  > p₂  > p ₃