A photon and an electron, each of 20 eV energy, move in free space. The ratio of linear momentum of electron ($p_e$) to that of photon ($p_{Ph}$), $p_e/p_{Ph}$, is: [Take speed of light = $3 imes10^8$ m s⁻¹, charge of electron = $-1.6 imes10^{-19}$ C and mass of electron = $9 imes10^{-31}$ kg]

Question

A photon and an electron, each of 20 eV energy, move in free space. The ratio of linear momentum of electron ($p_e$) to that of photon ($p_{Ph}$), $p_e/p_{Ph}$, is: [Take speed of light = $3	imes10^8$ m s⁻¹, charge of electron = $-1.6	imes10^{-19}$ C and mass of electron = $9	imes10^{-31}$ kg]

Dr. Abubacker · Accepted Answer

[Solution from source] E = 20 eV = $3.2	imes10^{-18}$ J. Photon: $p=E/c\approx1.07	imes10^{-26}$ kg m s⁻¹. Electron: $p=\sqrt{2mE}=2.4	imes10^{-24}$ kg m s⁻¹. Ratio = 225. Answer (4).

Explanation

Reviewed by Dr. Abubacker

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