Question

(a) Using the relativistic relation between E and p, show that electrons and photons with the same energy E have different wavelengths. (Note: Even at relativistic energies, the de Broglie wavelength equation is valid.) (b) Show that the ratio of their wavelengths approach equality as their common energy E gets much larger than mec^.

Answer 1

Since, we know that E^2 = P^2 c^2 + m^2_0 c^4 E^2 - m^2_0 c^4/c^2 = P^2 P = squareroot (E/c)^2 - m^2_0 c^2 So wavelength (lambda) = h/P lambda = h/(E/c)^2 - m^2_0 c^2 for me same E for and photons the lambda is Different, because rest mass (m_0) of (m_0)_e notequalto (m_0_photon) This lambda_e notequalto lambda_photon \r\n Since lambda_e = h/(E/c)^2 - m^2_e c^2 lambda_photon = h/(E/c)^2 - m^2_ph c^2 if E > > m_e c^2 (E/c)^2 - m^2_e c^2 tildeequal (E/c)^2 (E/c)^2 - m^2_ph c^2 tildeequal (E/c)^2 Therefore, lambda = h/squareroot (E/c)^2 = C h/E lambda_photon = h/squareroot (E/c) = Ch/E lambda_e = lambda_photon