The de Broglie wavelength of an electron moving with a velocity 1.5 × 108 m/s is equal to that of a photon. The ratio of the energy of the photon to the kinetic energy of the electron is ______.
The de Broglie wavelength (λ) for both the electron and the photon is given by: λ = h/p, where h is Planck's constant (h = 6.63 × 10⁻³⁴ J·s). The momentum of the electron (pₑ) is mₑv with mₑ = 9.11 × 10⁻³¹ kg and v = 1.5 × 10⁸ m/s. The momentum of the photon (pₚ) is Eₚ/c, where Eₚ is the energy of the photon and c is the speed of light (c = 3 × 10⁸ m/s). Since λ is equal for both the electron and the photon: h/mₑv = h/Eₚ/c. Simplifying gives Eₚ = mₑvc. The kinetic energy (Kₑ) of the electron is (1/2)mₑv². The ratio of the photon energy to the electron's kinetic energy is: Eₚ/Kₑ = (mₑvc)/((1/2)mₑv²)=2c/v=2×3×10⁸/1.5×10⁸=4. Thus, the ratio is 4, which is within the given range of 4,4.
