The wavelength λ of a photon and the de Broglie wavelength of an electron of massm have the same value. Show that the energy of the photon is 2λmc/h times thekinetic energy of the electron, where c and h have their usual meanings.

Sure, let's break this down step by step:

1. The energy of a photon is given by the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength.

2. The kinetic energy of an electron is given by the equation KE = 1/2 mv², where m is the mass of the electron and v is its velocity.

3. The de Broglie wavelength of a particle is given by the equation λ = h/p, where p is the momentum of the particle. For an electron, p = mv, so we can write λ = h/mv.

4. Setting the photon wavelength equal to the de Broglie wavelength of the electron gives us hc/λ = h/mv. Solving for v gives us v = mc/h.

5. Substituting this into the kinetic energy equation gives us KE = 1/2 m(mc/h)² = mc²/2h².

6. The ratio of the photon energy to the electron kinetic energy is then E/KE = (hc/λ) / (mc²/2h²) = 2λmc/h.

So, the energy of the photon is indeed 2λmc/h times the kinetic energy of the electron.