An electron and a photon initially have the same de-Broglie wavelength ₀ . Let K_e and E_ ph be their respective kinetic energy and total energy. If the de-Broglie wavelength of both particles is then changed to ₀ 2 , the ratio of their energies ( K_e E_ ph ) becomes n times its initial value. The value of n is:

Step-by-step solution

The kinetic energy of the electron in terms of its de-Broglie wavelength is: K_e = p^2 2m = h^2 2m ^2 The energy of the photon in terms of its wavelength is: E_ ph = hc The ratio of their energies is: R = K_e E_ ph = h^2 2m ^2 hc = h 2mc This shows that the energy ratio R is inversely proportional to the wavelength ( R 1 ). When the wavelength is halved ( ' = ₀ 2 ), the new ratio becomes: R' = h 2mc ( ₀ 2 ) = 2 ( h 2mc ₀ ) = 2R Therefore, the ratio becomes 2 times its initial value. Answer: 2